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a/ đk: \(\left[{}\begin{matrix}x\le\frac{-5-3\sqrt{5}}{10}\\x\ge\frac{-5+3\sqrt{5}}{10}\end{matrix}\right.\)\(\sqrt{x^2+x+1}+\sqrt{3x^2+3x+2}=\sqrt{5x^2+5x-1}\)
\(\Leftrightarrow\sqrt{x^2+x+1}+\sqrt{3\left(x^2+x+1\right)-1}=\sqrt{5\left(x^2+x+1\right)-6}\)
đặt\(x^2+x+1=t\left(t>0\right)\)
\(\sqrt{t}+\sqrt{3t-1}=\sqrt{5t-6}\)
bình phương 2 vế pt trở thành:
\(t+3t-1+2\sqrt{t\left(3t-1\right)}=5t-6\)
\(\Leftrightarrow2\sqrt{3t^2-t}=t-5\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left(2\sqrt{3t^2-t}\right)^2=\left(t-5\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\11t^2+6t-25=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}t\ge5\\\left[{}\begin{matrix}t=\frac{-3+2\sqrt{71}}{11}\\t=\frac{-3-2\sqrt{71}}{11}\end{matrix}\right.\end{matrix}\right.\)=> không có gtri t nào t/m
vậy pt vô nghiệm
a/ ĐKXĐ: ...
Đặt \(x^2+x+1=a>0\)
\(\sqrt{a}+\sqrt{3a-1}=\sqrt{5a-6}\)
\(\Leftrightarrow4a-1+2\sqrt{3a^2-a}=5a-6\)
\(\Leftrightarrow2\sqrt{3a^2-a}=a-5\) (\(a\ge5\))
\(\Leftrightarrow4\left(3a^2-a\right)=a^2-10a+25\)
\(\Leftrightarrow11a^2+6a-25=0\)
Nghiệm xấu quá, chắc bạn nhầm lẫn đâu đó
b/
Đặt \(x^2+x+1=a>0\)
\(\sqrt{a+3}+\sqrt{a}=\sqrt{2a+7}\)
\(\Leftrightarrow2a+3+2\sqrt{a^2+3a}=2a+7\)
\(\Leftrightarrow\sqrt{a^2+3a}=2\)
\(\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x^2+x+1=1\)
\(\Leftrightarrow2\left(x^2+1\right)-2x\sqrt{x^2+1}=5\)
\(\Leftrightarrow x^2+1-2x\sqrt{x^2+1}+x^2=4\)
\(\Leftrightarrow\left(\sqrt{x^2+1}-x\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+1}-x=2\\\sqrt{x^2+1}-x=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+1}=x+2\left(x\ge-2\right)\\\sqrt{x^2+1}=x-2\left(x\ge2\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2+4x+4\\x^2+1=x^2-4x+4\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\frac{3}{4}\\x=\frac{3}{4}< 2\left(l\right)\end{matrix}\right.\)
ĐKXĐ: \(x\ge\frac{3}{2}\)
\(\Leftrightarrow\sqrt{5x-1}+\sqrt{2x-3}=\sqrt{3x-2}\)
\(\Leftrightarrow7x-4+2\sqrt{\left(5x-1\right)\left(2x-3\right)}=3x-2\)
\(\Leftrightarrow\sqrt{10x^2-17x+3}=1-2x\)
Do \(x\ge\frac{3}{2}\Rightarrow1-2x< 0\)
Phương trình vô nghiệm
Lời giải:
ĐKXĐ: \(x^2+5x+2\geq 0\)
PT \(\Leftrightarrow x^2+5x+4-3\sqrt{x^2+5x+2}-6=0\)
\(\Leftrightarrow (x^2+5x+2)-3\sqrt{x^2+5x+2}-4=0\)
\(\Leftrightarrow a^2-3a-4=0\) (đặt \(a=\sqrt{x^2+5x+2}; a\geq 0)\)
\(\Leftrightarrow (a-4)(a+1)=0\Rightarrow a=4\) (do $a\geq 0)$
\(\Rightarrow x^2+5x+2=16\)
\(\Leftrightarrow x^2+5x-14=0\)
\(\Leftrightarrow (x-2)(x+7)=0\Rightarrow \left[\begin{matrix} x=2\\ x=-7\end{matrix}\right.\) (đều thỏa mãn)
\(pt\Leftrightarrow2\left(x+1\right)\sqrt{x}+\sqrt{3\left(2x+1\right)\left(x+1\right)^2}=\left(x+1\right)\left(5x^2-8x+8\right)\)\(\Leftrightarrow2\left(x+1\right)\sqrt{x}+\left(x+1\right)\sqrt{3\left(2x+1\right)}-\left(x+1\right)\left(5x^2-8x+8\right)=0\)\(\Leftrightarrow\left(x+1\right)\left(2\sqrt{x}+\sqrt{3\left(2x+1\right)}-5x^2+8x-8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\2\sqrt{x}+\sqrt{3\left(2x+1\right)}-5x^2-8+8x=0\circledast\end{matrix}\right.\)
Giải (*)\(2\sqrt{x}+\sqrt{3\left(2x+1\right)}-5x^2-8+8x=0\)
\(\Leftrightarrow2\sqrt{x}-2+\sqrt{3\left(2x+1\right)}-3=5x^2-8x+3\)
\(\Leftrightarrow\frac{4x-4}{2\sqrt{x}+2}+\frac{6x-6}{\sqrt{3\left(2x+1\right)}+3}=\left(x-1\right)\left(5x-3\right)\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{2}{\sqrt{x}+1}+\frac{6}{\sqrt{3\left(2x+1\right)}+3}-5x+3\right)=0\)
x=1
bạn giải nốt cái còn lại nhá
e/
ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow x^2+8x-2+6\sqrt{x\left(x+1\right)\left(x-2\right)}\le5x^2-4x-6\)
\(\Leftrightarrow3\sqrt{x\left(x+1\right)\left(x-2\right)}\le2x^2-6x-2\)
\(\Leftrightarrow3\sqrt{\left(x^2-2x\right)\left(x+1\right)}\le2x^2-6x-2\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-2x}=a\ge0\\\sqrt{x+1}=b>0\end{matrix}\right.\)
\(\Rightarrow2a^2-2b^2=2x^2-6x-2\)
BPT trở thành:
\(3ab\le2a^2-2b^2\Leftrightarrow2a^2-3ab-2b^2\ge0\)
\(\Leftrightarrow\left(2a+b\right)\left(a-2b\right)\ge0\)
\(\Leftrightarrow a\ge2b\Rightarrow\sqrt{x^2-2x}\ge2\sqrt{x+1}\)
\(\Leftrightarrow x^2-2x\ge4x+4\)
\(\Leftrightarrow x^2-6x-4\ge0\)
\(\Rightarrow x\ge3+\sqrt{13}\)
d/
ĐKXĐ: \(x\ge-1\)
\(3\sqrt{\left(x+1\right)\left(x^2-x+1\right)}+4x^2-5x+3\ge0\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-x+1}=a>0\\\sqrt{x+1}=b\ge0\end{matrix}\right.\)
\(\Rightarrow4a^2-b^2=4x^2-5x+3\)
BPT trở thành:
\(4a^2+3ab-b^2\ge0\)
\(\Leftrightarrow\left(a+b\right)\left(4a-b\right)\ge0\)
\(\Leftrightarrow4a-b\ge0\Rightarrow4a\ge b\)
\(\Rightarrow4\sqrt{x^2+x+1}\ge\sqrt{x+1}\)
\(\Leftrightarrow16x^2+16x+4\ge x+1\)
\(\Leftrightarrow16x^2+15x+3\ge0\)
\(\Rightarrow\left[{}\begin{matrix}-1\le x\le\frac{-15-\sqrt{33}}{32}\\x\ge\frac{-15+\sqrt{33}}{32}\end{matrix}\right.\)
a/ ĐKXĐ: \(-\frac{1}{2}\le x\le4\)
\(\sqrt{4-x}=\sqrt{x+1}+\sqrt{2x+1}\)
\(\Leftrightarrow4-x=3x+2+2\sqrt{2x^2+3x+1}\)
\(\Leftrightarrow1-2x=\sqrt{2x^2+3x+1}\) (\(x\le\frac{1}{2}\))
\(\Leftrightarrow4x^2-4x+1=2x^2+3x+1\)
\(\Leftrightarrow2x^2-7x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{7}{2}\left(l\right)\end{matrix}\right.\)
Bài này liên hợp cũng được
b/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{5x+1}^2-\sqrt{5x+1}\left(\sqrt{14x+7}-\sqrt{2x+3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\Rightarrow x=-\frac{1}{5}\\\sqrt{5x+1}-\sqrt{14x+7}+\sqrt{2x+3}=0\left(1\right)\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow\sqrt{5x+1}+\sqrt{2x+3}=\sqrt{14x+7}\)
\(\Leftrightarrow7x+4+2\sqrt{10x^2+17x+3}=14x+7\)
\(\Leftrightarrow2\sqrt{10x^2+17x+3}=7x+3\)
\(\Leftrightarrow4\left(10x^2+17x+3\right)=\left(7x+3\right)^2\)
\(\Leftrightarrow...\)
c/ ĐKXĐ: \(x\ge\frac{1}{2}\)
Đặt \(\left\{{}\begin{matrix}\sqrt[3]{2-2x}=a\\\sqrt{2x-1}=b\end{matrix}\right.\) ta được:
\(\left\{{}\begin{matrix}a=1-b\\a^3+b^2=1\end{matrix}\right.\) \(\Rightarrow a^3+\left(1-a\right)^2=1\)
\(\Leftrightarrow a^3+a^2-2a=0\)
\(\Leftrightarrow a\left(a^2+a-2\right)=0\Rightarrow\left[{}\begin{matrix}a=0\\a=1\\a=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2-2x=0\\2-2x=1\\2-2x=-8\end{matrix}\right.\)
d/ ĐKXĐ: \(x\le\frac{5}{4}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{5-4x}=a\\\sqrt[3]{x+7}=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=3\\a^2+4b^3=33\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=3-b\\a^2+4b^3=33\end{matrix}\right.\)
\(\Leftrightarrow\left(3-b\right)^2+4b^3=33\)
\(\Leftrightarrow4b^3+b^2-6b-24=0\)
\(\Leftrightarrow\left(b-2\right)\left(4b^2+9b+12\right)=0\)
\(\Rightarrow b=2\Rightarrow\sqrt[3]{x+7}=2\Rightarrow x=1\)
a: ĐKXĐ: \(\left(2x^2-5x+2\right)\left(x^3+1\right)< >0\)
=>(2x-1)(x-2)(x+1)<>0
hay \(x\notin\left\{\dfrac{1}{2};2;-1\right\}\)
b: ĐKXĐ: x+5<>0
=>x<>-5
c: ĐKXĐ: x4-1<>0
hay \(x\notin\left\{1;-1\right\}\)
d: ĐKXĐ: \(x^4+2x^2-3< >0\)
=>\(x\notin\left\{1;-1\right\}\)
a/ \(\Leftrightarrow\sqrt{x^2+x+3}-\sqrt{x^2+2}+\sqrt{x^2+x+8}-\sqrt{x^2+7}=0\)
\(\Leftrightarrow\frac{x+1}{\sqrt{x^2+x+3}+\sqrt{x^2+2}}+\frac{x+1}{\sqrt{x^2+x+8}+\sqrt{x^2+7}}=0\)
\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{\sqrt{x^2+x+3}+\sqrt{x^2+2}}+\frac{1}{\sqrt{x^2+x+8}+\sqrt{x^2+7}}\right)=0\)
\(\Leftrightarrow x+1=0\) (ngoặc to phía sau luôn dương)
\(\Rightarrow x=-1\)
b/
\(\sqrt{7-x^2+x\sqrt{x+5}}=\sqrt{3-2x-x^2}\) (1)
\(\Rightarrow7-x^2+x\sqrt{x+5}=3-2x-x^2\)
\(\Leftrightarrow x\sqrt{x+5}=-2x-4\)
\(\Rightarrow x^2\left(x+5\right)=4x^2+16x+16\)
\(\Rightarrow x^3+x^2-16\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=2\\x=-2\end{matrix}\right.\)
Do các phép biến đổi ko tương đương nên cần thay nghiệm vào (1) để kiểm tra
c/ ĐKXĐ: \(x\ge\frac{5}{3}\)
\(\Leftrightarrow\sqrt{10x+1}-\sqrt{9x+4}+\sqrt{3x-5}-\sqrt{2x-2}=0\)
\(\Leftrightarrow\frac{x-3}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{x-3}{\sqrt{3x-5}+\sqrt{2x-2}}=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{1}{\sqrt{10x+1}+\sqrt{9x+4}}+\frac{1}{\sqrt{3x-5}+\sqrt{2x-2}}\right)=0\)
\(\Leftrightarrow x-3=0\) (ngoặc phía sau luôn dương)
d/ Đề bài là \(2\sqrt{2x+3}\) hay \(2\sqrt{2x-3}\) bạn?
e/ ĐKXĐ: \(x\ge-3\)
\(\Leftrightarrow\sqrt{x+3+2\sqrt{x+3}+1}=x+4\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+3}+1\right)^2}=x+4\)
\(\Leftrightarrow\sqrt{x+3}+1=x+4\)
\(\Leftrightarrow x+3-\sqrt{x+3}=0\)
\(\Leftrightarrow\sqrt{x+3}\left(\sqrt{x+3}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x+3=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
ĐKXĐ: ...
\(\sqrt[3]{x^3+5x^2}-x-2+x+1-\sqrt{\dfrac{5x^2-2}{6}}=0\)
\(\Leftrightarrow\dfrac{x^3+5x^2-\left(x+2\right)^3}{\left(x+2\right)^2+\left(x+2\right)\sqrt[3]{x^3+5x^2}+\sqrt[3]{\left(x^3+5x^2\right)^2}}+\dfrac{\left(x+1\right)^2-\dfrac{5x^2-2}{6}}{x+1+\sqrt{\dfrac{5x^2-2}{6}}}=0\)
\(\Leftrightarrow\left(x^2+12x+8\right)\left(\dfrac{1}{6\left(x+1\right)+\sqrt{6\left(5x^2-2\right)}}-\dfrac{1}{\left(x+2\right)^2+\left(x+2\right)\sqrt[3]{x^3+5x^2}+\sqrt[3]{\left(x^3+5x^2\right)^2}}\right)=0\)
\(\Leftrightarrow x^2+12x+8=0\)