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a/ ĐKXĐ: ...
\(\Leftrightarrow2\left(x^2-5x-6\right)+\sqrt{x^2-5x-6}-3=0\)
Đặt \(\sqrt{x^2-5x-6}=a\ge0\)
\(2a^2+a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-5x-6}=1\Leftrightarrow x^2-5x-7=0\)
b/ ĐKXĐ: ...
\(\Leftrightarrow5\sqrt{3x^2-4x-2}-2\left(3x^2-4x-2\right)+3=0\)
Đặt \(\sqrt{3x^2-4x-2}=a\ge0\)
\(-2a^2+5a+3=0\) \(\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{3x^2-4x-2}=3\Leftrightarrow3x^2-4x-11=0\)
c/ \(\Leftrightarrow x^2+2x-6+\sqrt{2x^2+4x+3}=0\)
Đặt \(\sqrt{2x^2+4x+3}=a>0\Rightarrow x^2+2x=\frac{a^2-3}{2}\)
\(\frac{a^2-3}{2}-6+a=0\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-5\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{2x^2+4x+3}=3\Leftrightarrow2x^2+4x-6=0\)
d/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{3x-1}{x}}=a>0\)
\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\)
\(\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)
\(\Rightarrow a=1\Rightarrow\sqrt{\frac{3x-1}{x}}=1\Leftrightarrow3x-1=x\)
e/ĐKXĐ: ...
\(\Leftrightarrow2\sqrt{\frac{6x-1}{x}}=\frac{x}{6x-1}+1\)
Đặt \(\sqrt{\frac{6x-1}{x}}=a>0\)
\(2a=\frac{1}{a^2}+1\Leftrightarrow2a^3-a^2-1=0\Leftrightarrow\left(a-1\right)\left(2a^2+a+1\right)=0\)
\(\Rightarrow a=1\Rightarrow\sqrt{\frac{6x-1}{x}}=1\Rightarrow6x-1=x\)
f/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{x}{2x-1}}=a>0\)
\(\frac{1}{a}+1+a=3a^2\)
\(\Leftrightarrow3a^3-a^2-a-1=0\)
\(\Leftrightarrow\left(a-1\right)\left(3a^2+2a+1\right)=0\)
\(\Leftrightarrow a=1\Rightarrow\sqrt{\frac{x}{2x-1}}=1\Rightarrow x=2x-1\)
a/ \(x\ge-3\)
\(\Leftrightarrow\left(2x-1\right)^2=\left(x+3\right)^2\)
\(\Leftrightarrow3x^2-10x-8=0\Rightarrow\left[{}\begin{matrix}x=4\\x=-\frac{2}{3}\end{matrix}\right.\)
b/ \(x\ge-\frac{5}{2}\)
\(\Leftrightarrow\left(4x+7\right)^2=\left(2x+5\right)^2\)
\(\Leftrightarrow x^2+3x+2=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
c/ \(x\ge1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-5=5x-5\\2x^2-3x-5=5-5x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2-8x=0\\2x^2+2x-10=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=4\\x=\frac{-1+\sqrt{21}}{2}\\x=\frac{-1-\sqrt{21}}{2}\left(l\right)\end{matrix}\right.\)
d/ \(x\ge\frac{17}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=4x-17\\x^2-4x-5=17-4x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-8x+12=0\\x^2=22\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\left(l\right)\\x=\sqrt{22}\\x=-\sqrt{22}\left(l\right)\end{matrix}\right.\)
e/ \(\left[{}\begin{matrix}x\ge1\\x\le-\frac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-x-2=x-2\\3x^2-x-2=2-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=0\\3x^2=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=\frac{2}{3}\left(l\right)\\x=\frac{2\sqrt{3}}{3}\\x=\frac{-2\sqrt{3}}{3}\end{matrix}\right.\)
đa phần mình sử dụng phương pháp liên hợp nha bạn
\(\sqrt{a}-\sqrt{b}=\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\)
b. điều kiện \(\dfrac{1}{4}\le x\le\dfrac{3}{8}\), pt:
\(\Leftrightarrow\sqrt{3-8x}-\sqrt{4x-1}=6x-2\\ \Leftrightarrow\dfrac{3-8x-4x+1}{\sqrt{3-8x}+\sqrt{4x-1}}=2\left(3x-1\right)\\ \Leftrightarrow\dfrac{-4\left(3x-1\right)}{\sqrt{3-8x}+\sqrt{4x-1}}=2\left(3x-1\right)\\ \Leftrightarrow2\left(3x-1\right)+\dfrac{4\left(3x-1\right)}{\sqrt{3-8x}+\sqrt{4x-1}}=0\\ \Leftrightarrow2\left(3x-1\right)\left(1+\dfrac{2}{\sqrt{3-8x}+\sqrt{4x-1}}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(n\right)\\1+\dfrac{2}{\sqrt{3-8x}+\sqrt{4x-1}}=0\left(vn\right)\end{matrix}\right.\)
d. điều kiện: \(x\le-4\cup x\ge0\), pt:
\(\Leftrightarrow1-\sqrt{x^2-3x+3}=\sqrt{2x^2+x+2}-\sqrt{x^2+4x}\\ \Leftrightarrow\dfrac{1-x^2+3x-3}{1+\sqrt{x^2-3x+3}}=\dfrac{2x^2+x+2-x^2-4x}{\sqrt{2x^2+x+2}+\sqrt{x^2+4x}}\\ \Leftrightarrow\dfrac{-\left(x-1\right)\left(x-2\right)}{1+\sqrt{x^2-3x+3}}=\dfrac{\left(x-1\right)\left(x-2\right)}{\sqrt{2x^2+x+2}+\sqrt{x^2+4x}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=1\left(n\right)\\\dfrac{-1}{1+\sqrt{x^2-3x+3}}=\dfrac{1}{\sqrt{2x^2+x+2}+\sqrt{x^2+4x}}\left(vn\right)\end{matrix}\right.\)
e. điều kiện:x thuộc R
\(\Leftrightarrow\sqrt{x^2+15}-4=3x-3+\sqrt{x^2+8}-3\\ \Leftrightarrow\dfrac{x^2+15-16}{\sqrt{x^2+15}+4}=3\left(x-1\right)+\dfrac{x^2+8-9}{\sqrt{x^2+8}+3}\\ \Leftrightarrow\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+15}+4}-3\left(x-1\right)-\dfrac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+8}+3}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\\dfrac{\left(x+1\right)}{\sqrt{x^2+15}+4}-3-\dfrac{\left(x+1\right)}{\sqrt{x^2+8}+3}=0\left(1\right)\end{matrix}\right.\)
(1) mình không biết có vô nghiệm không nữa và cũng thua luôn
f. điều kiện: \(x\ge-2\)
bài này giải cách hơi khác một chút
đặt \(a=\sqrt{x+5}\left(\ge0\right)\\ b=\sqrt{x+2}\left(\ge0\right)\)
pt:
\(\Leftrightarrow\left(\sqrt{x+5}-\sqrt{x+2}\right)\left[\left(1+\sqrt{\left(x+5\right)\left(x+2\right)}\right)\right]\\ \Rightarrow\left(a-b\right)\left(1+ab\right)=3\left(1\right)\)
mà \(a^2-b^2=x+5-x-2=3\\ \Rightarrow\left(a-b\right)\left(a+b\right)=3\left(2\right)\)
=> (1) = (2)
\(\Leftrightarrow\left(a-b\right)\left(1+ab\right)=\left(a-b\right)\left(a+b\right)\\ \Leftrightarrow\left(a-b\right)\left(1+ab-a-b\right)=0\\ \Leftrightarrow\left(a-b\right)\left(a-1\right)\left(b-1\right)=0\)
TH1: a=b \(\Leftrightarrow\sqrt{x+5}=\sqrt{x+2}\Leftrightarrow x+5=x+2\left(vn\right)\)
TH2: a=1\(\Leftrightarrow\sqrt{x+5}=1\Leftrightarrow x=-4\left(l\right)\)
TH3: b=1\(\Leftrightarrow\sqrt{x+2}=1\Leftrightarrow x=-1\left(n\right)\)
g. điều kiện: \(x\le-\sqrt{2}\cup x\ge\dfrac{7+\sqrt{37}}{2}\)
pt:
\(\dfrac{3x^2-7x+3-3x^2+5x+1}{\sqrt{3x^2-7x+2}+\sqrt{x^2-3x-4}}=\dfrac{x^2-2-x^2+3x-4}{\sqrt{3x^2-5x-1}+\sqrt{x^2-2}}\\ \Leftrightarrow\dfrac{-2\left(x-2\right)}{\sqrt{3x^2-7x+2}+\sqrt{x^2-3x-4}}=\dfrac{3\left(x-2\right)}{\sqrt{3x^2-5x-1}+\sqrt{x^2-2}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\left(n\right)\\\dfrac{-2}{\sqrt{3x^2-7x+2}+\sqrt{x^2-3x-4}}=\dfrac{3}{\sqrt{3x^2-5x-1}+\sqrt{x^2-2}}\left(vn\right)\end{matrix}\right.\)h. điều kiện \(x\le-2-\sqrt{7}\cup x\ge-2+\sqrt{7}\)
\(\sqrt{2x^2+x-1}-\sqrt{x^2+4x-3}=\sqrt{2x^2+4x-3}-\sqrt{3x^2+x-1}\\ \Leftrightarrow\dfrac{2x^2+x-1-x^2-4x+3}{\sqrt{2x^2+x-1}+\sqrt{x^2+4x-3}}=\dfrac{2x^2+4x-3-3x^2-x+1}{\sqrt{2x^2+4x-3}+\sqrt{3x^2+x-1}}\\ \Leftrightarrow\dfrac{x^2-3x+2}{\sqrt{2x^2+x-1}+\sqrt{x^2+4x-3}}=\dfrac{-\left(x^2-3x+2\right)}{\sqrt{2x^2+4x-3}+\sqrt{3x^2+x-1}}\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+2=0\Leftrightarrow x=1\left(n\right),x=2\left(n\right)\\\dfrac{1}{\sqrt{2x^2+x-1}+\sqrt{x^2+4x-3}}=\dfrac{-1}{\sqrt{2x^2+4x-3}+\sqrt{3x^2+x-1}}\left(vn\right)\end{matrix}\right.\)
(nhớ tích cho mình nha, mấy bài kia mình ko biết làm huhu)
\(1,\left|2x-3\right|=x-5\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-5\ge0\\\left[{}\begin{matrix}2x-3=x-5\\2x-3=-x+5\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}5\\\left[{}\begin{matrix}x=-2\\x=\frac{8}{3}\end{matrix}\right.\end{matrix}\right.\) (ko thỏa mãn)
=> pt vô nghiệm
\(2,\left|3x+2\right|=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1\text{≥}0\\\left[{}\begin{matrix}3x+2=x+1\\3x+2=-x-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}-1\\\left[{}\begin{matrix}x=-\frac{1}{2}\\x=-\frac{3}{4}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\\x=-\frac{3}{4}\end{matrix}\right.\)
\(3,\left|2x+1\right|=7-x\)
\(\Leftrightarrow\left\{{}\begin{matrix}7-x\text{≥}0\\\left[{}\begin{matrix}2x+1=7-x\\2x+1=x-7\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}7\\\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\end{matrix}\right.\) (loại)
=> pt vô nghiệm
\(4,\left|2x-5\right|=x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1\text{≥}0\\\left[{}\begin{matrix}2x-5=x+1\\2x-5=-x-1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}-1\\\left[{}\begin{matrix}x=6\\x=\frac{4}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{4}{3}\end{matrix}\right.\)
\(5,\left|6x-2\right|=3x-4\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-4\text{≥}0\\\left[{}\begin{matrix}6x-2=3x-4\\6x-2=-3x+4\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}\frac{4}{3}\\\left[{}\begin{matrix}x=-\frac{2}{3}\\x=\frac{2}{3}\end{matrix}\right.\end{matrix}\right.\) => pt vô nghiệm
\(6,\left|3x-2\right|=x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2\text{≥}0\\\left[{}\begin{matrix}3x-2=x-2\\3x-2=-x+2\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\text{≥}2\\\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\end{matrix}\right.\) => pt vô nghiệm
\(7,\left|2x+3\right|=1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=1\\2x+3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)
\(8,\left|2-x\right|=2x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-1\ge0\\\left[{}\begin{matrix}2-x=2x-1\\2-x=-2x+1\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{1}{2}\\\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow x=1\)
\(9,\left|2x-1\right|=x-3\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3\ge0\\\left[{}\begin{matrix}2x-1=x-3\\2x-1=-x+3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge3\\\left[{}\begin{matrix}x=-2\\x=\frac{4}{3}\end{matrix}\right.\end{matrix}\right.\) => pt vô nghiệm
\(10,2\left|x-1\right|=x+2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2\ge0\\\left[{}\begin{matrix}2x-2=x+2\\2x-2=-x-2\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-2\\\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)
a)
\(\left\{{}\begin{matrix}x^2+x+5< 0\\x^2-6x+1>0\end{matrix}\right.\)
\(\)Ta có
\(x^2+x+5=\left(x^2+x+\dfrac{1}{4}\right)+\dfrac{19}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{19}{4}\ge\dfrac{19}{4}>0\)
=> Bất phương trình đàu tiên sai, hệ bất phương trình sai
b)
\(\left\{{}\begin{matrix}2x^2+x-6>0\\3x^2-10x+3\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-3\right)\left(x+2\right)>0\\\left(x-3\right)\left(3x-1\right)\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>2\\x< -3\end{matrix}\right.\\\left[{}\begin{matrix}x\le-\dfrac{1}{3}\\x\ge3\end{matrix}\right.\end{matrix}\right.\)
a)
\(\left|x+7\right|=2x+5\\ \Rightarrow\left(x+7\right)^2=\left(2x+5\right)^2\\ \left(x+7\right)^2-\left(2x+5\right)^2=0\\ \left(x+7+2x+5\right)\left(x+7-2x-5\right)=0\\ \left(3x+12\right)\left(2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+12=0\\2-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)
vậy phương trình đã cho có tập nghiệm là S={-4;2}
b)
\(\left|3x-4\right|-2x=1\\ \left|3x-4\right|=1+2x\\ \left(3x-4\right)^2-\left(1+2x\right)^2=0\\ \left(3x-4+1+2x\right)\left(3x-4-1-2x\right)=0\\ \left(5x-3\right)\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}5x-3=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=5\end{matrix}\right.\)
vậy phương trình đã cho có tập nghiệm là S={3/5;5}
c)
\(5.\left|1-4x\right|=10x-5\\ \left|1-4x\right|=\dfrac{10x-5}{5}=2x-1\\ \Rightarrow\left(1-4x\right)^2-\left(2x-1\right)^2=0\\ \left(1-4x+2x-1\right)\left(1-4x-2x+1\right)=0\\ -2x\left(-6x+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\-6x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
vậy phương trình đã cho có tập nghiệm là S={0;3}
d)
\(\left|2x+3\right|=12-6x\\ \left(2x+3\right)^2-\left(12-6x\right)^2=0\\ \left(2x+3+12-6x\right)\left(2x+3+6x-12\right)=0\\ \left(15-4x\right)\left(8x-9\right)=0\\ \Rightarrow\left[{}\begin{matrix}15-4x=0\\8x-9=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{4}\\x=\dfrac{9}{8}\end{matrix}\right.\)
vậy phương trình đã cho có tập nghiệm là S={15/4;9/8}
a) \(\left|x+7\right|=2x+5\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+7=2x+5\\x+7=-2x-5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
b)\(\left|3x-4\right|=2x+1\)
\(\Leftrightarrow\left\{{}\begin{matrix}3x-4=2x+1\\3x-4=-2x-1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=5\\x=\dfrac{3}{5}\end{matrix}\right.\)
c)\(\left|1-4x\right|=2x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}1-4x=2x-1\\1-4x=1-2x\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x=0\end{matrix}\right.\)
d)\(\left|2x+3\right|=12-6x\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3=12-6x\\2x+3=6x-12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{9}{8}\\x=\dfrac{15}{4}\end{matrix}\right.\)