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MK mới học lớp 8 thui nên làm được bài 4
Câu 4:
a)\(3x^2-4x+1\)
\(\Leftrightarrow3x^2-3x-x+1\)
\(\Leftrightarrow3x\left(x-1\right)-\left(x-1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x-1\right)\)
b)\(x^2-y^2+4x+4\)
\(\Leftrightarrow x^2+4x+4-y^2\)
\(\Leftrightarrow\left(x+2\right)^2-y^2\)
\(\Leftrightarrow\left(x+2-y\right)\left(x+2+y\right)\)
a: \(=\left|x-4\right|-\left|x-2\right|\)
\(=\left|3\sqrt{2}-1-4\right|-\left|3\sqrt{2}-1-2\right|\)
\(=5-3\sqrt{2}-\left(3\sqrt{2}-3\right)=-6\sqrt{2}+8\)
b: \(=\left|\sqrt{x-1}+1\right|+\left|\sqrt{x-1}-1\right|\)
\(=\left|\sqrt{7}-1+1\right|+\left|\sqrt{7}-1-1\right|\)
\(=\sqrt{7}+4-\sqrt{7}=4\)
a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+6\right)^2=\left(1-x\right)^2\\-3< =x< =1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(2x+6+x-1\right)\left(2x+6+1-x\right)=0\\-3< =x< =1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x+5\right)\left(x+7\right)=0\\-3< =x< =1\end{matrix}\right.\Leftrightarrow x=-\dfrac{5}{3}\)
b: \(\Leftrightarrow2\cdot3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=2x\)
\(\Leftrightarrow4\sqrt{x-3}=2x\)
\(\Leftrightarrow2\sqrt{x-3}=x\)
\(\Leftrightarrow\sqrt{4x-12}=x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=3\\x^2=4x-12\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
a) \(\sqrt{x-3}\) xác định
\(\Leftrightarrow x-3\ge0\)
\(\Leftrightarrow x\ge3\)
Vậy..
b) \(\sqrt{3-2x}\) xác định
\(\Leftrightarrow3-2x\ge0\)
\(\Leftrightarrow x\le-\dfrac{3}{2}\)
Vậy..
c) \(\sqrt{4x^2-1}\) xác định
\(\Leftrightarrow4x^2-1\ge0\)
\(\Leftrightarrow\left(2x-1\right)\left(2x+1\right)\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}2x-1\ge0\\2x+1\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge\dfrac{1}{2}\\x\ge\dfrac{-1}{2}\end{matrix}\right.\)\(\Rightarrow x\ge\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}2x-1\le0\\2x+1\le0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\x\le\dfrac{-1}{2}\end{matrix}\right.\) \(\Rightarrow x\le\dfrac{-1}{2}\)
Vậy ...
d) \(\sqrt{3x^2+2}\) xác định
\(\Leftrightarrow3x^2+2\ge0\)
mà \(3x^2\ge0\)
\(\Rightarrow3x^2+2>0\)
Vậy...
e) \(\sqrt{2x^2+4x+5}\) xác định
\(\Leftrightarrow2x^2+4x+5\ge0\)
mà \(2x^2+4x\ge0\)
\(2x\left(x+2\right)\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}2x\ge0\\x+2\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ge0\\x\ge-2\end{matrix}\right.\)\(\Rightarrow x\ge0\)
\(\Rightarrow\left\{{}\begin{matrix}2x\le0\\x+2\le0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\le0\\x\le-2\end{matrix}\right.\)\(\Rightarrow x\le-2\)
\(\Rightarrow2x^2+4x+5>0\)
Vậy...
( Câu này không chắc lắm nha )
Bài 2: Tách sẵn ra cho bạn luôn nhé, không thì bạn nhấn máy tính ra cũng được :v
a) \(-\dfrac{7}{9}\sqrt{\left(-27\right)^2+6\sqrt{1}}\)
\(=-\dfrac{7}{9}\sqrt{\left(-3\right)^2.\left(-9\right)^2+6}\)
\(=\dfrac{-7}{9}\sqrt{735}\)
\(=\dfrac{-7}{9}\sqrt{49.15}\)
\(=\dfrac{-49\sqrt{15}}{9}\)
b) \(\sqrt{49}\sqrt{12^2}+\sqrt{256}:\sqrt{8^2}\)
\(=84+2=86\)
c)\(\sqrt{\left(\sqrt{3-1}\right)^2-\sqrt{\left(\sqrt{3+1}\right)^2}}\)
\(=\sqrt{2-2}\)
= 0
a: \(\Leftrightarrow\left\{{}\begin{matrix}\left(2x+6\right)^2=\left(1-x\right)^2\\-3< =x< =1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(2x+6+x-1\right)\left(2x+6+1-x\right)=0\\-3< =x< =1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(3x+5\right)\left(x+7\right)=0\\-3< =x< =1\end{matrix}\right.\Leftrightarrow x=-\dfrac{5}{3}\)
b: \(\Leftrightarrow2\cdot3\sqrt{x-3}-\dfrac{1}{5}\cdot5\sqrt{x-3}-\dfrac{1}{7}\cdot7\sqrt{x-3}=2x\)
\(\Leftrightarrow4\sqrt{x-3}=2x\)
\(\Leftrightarrow2\sqrt{x-3}=x\)
\(\Leftrightarrow\sqrt{4x-12}=x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=3\\x^2=4x-12\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
a) điều kiện : \(x>0;x\ne4\)
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{x-2\sqrt{x}}\right)\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{x-4}\right)\)
\(P=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right)\)
\(P=\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}-2+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}.\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(P=\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(x=4+2\sqrt{3}\Leftrightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}+1\right)^2}\Leftrightarrow\sqrt{x}=\sqrt{3}+1\) \(\left(x>0\right)\)
thay vào P ta có \(P=\dfrac{\sqrt{3}+1+2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}+1-2\right)}=\dfrac{\sqrt{3}+3}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{\sqrt{3}+3}{2}\)
\(P>0\Leftrightarrow\dfrac{\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-2\right)}>0\)
ta có : \(\sqrt{x}+2>0\) và \(\sqrt{x}>0\) \(\left(x>0\right)\)
\(\Rightarrow p>0\) thì \(\sqrt{x}-2>0\Leftrightarrow\sqrt{x}>2\Leftrightarrow x>4\)
vậy \(x>4\) thì P > 0
câu : a ; b ; c đầy đủ rồi nha quênh gi câu : a ; b ; c 
mình đang cần gấp lắm~!!!
a) \(\sqrt{\left|x\right|-1}\) biểu thức sau có nghĩa \(\Leftrightarrow\) \(\left|x\right|-1\ge0\)
\(\Leftrightarrow\left|x\right|\ge1\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge1\\hoac\\x\le-1\end{matrix}\right.\)
b) \(\sqrt{\left|x-1\right|-3}\) biểu thức sau có nghĩa \(\Leftrightarrow\left|x-1\right|-3\ge0\)
\(\Leftrightarrow\left|x-1\right|\ge3\) \(\left\{{}\begin{matrix}x-1\ge3\\hoac\\x-1\le-3\end{matrix}\right.\)
c) \(\sqrt{4-\left|x\right|}\) biểu thức sau có nghĩa \(\Leftrightarrow4-\left|x\right|\ge0\)
\(\Leftrightarrow4\ge\left|x\right|\) \(\Leftrightarrow-4\le x\le4\)
1, đk: \(x>0\) và \(x\ne4\)
Ta có: A=\(\dfrac{1}{2\sqrt{x}-x}=\dfrac{1}{-\left(x-2\sqrt{x}+1\right)+1}=\dfrac{1}{-\left(\sqrt{x}-1\right)^2+1}\)
Ta luôn có: \(-\left(\sqrt{x}-1\right)^2\le0\) với \(x>0\) và \(x\ne4\)
\(\Rightarrow-\left(\sqrt{x}-1\right)^2+1\le1\)
\(\Rightarrow A\ge1\). Dấu "=" xảy ra <=> x=1 (t/m)
Vậy MinA=1 khi x=1
2, đk: \(x\ge0;x\ne1;x\ne9\)
Ta có: B=\(\dfrac{1}{x-4\sqrt{x}+3}=\dfrac{1}{\left(x-4\sqrt{x}+4\right)-1}=\dfrac{1}{\left(\sqrt{x}-2\right)^2-1}\)
Ta luôn có: \(\left(\sqrt{x}-2\right)^2\ge0\) với \(x\ge0;x\ne1;x\ne9\)
\(\Rightarrow\left(\sqrt{x}-2\right)^2-1\ge-1\)
\(\Rightarrow B\le-1\). Dấu "=" xảy ra <=> x=4 (t/m)
Vậy MaxB=-1 khi x=4
3, đk: \(x\ge0;x\ne15+4\sqrt{11}\)
Ta có: C=\(\dfrac{1}{4\sqrt{x}-x+7}=\dfrac{1}{-\left(x-4\sqrt{x}+4\right)+11}=\dfrac{1}{-\left(\sqrt{x}-2\right)^2+11}\)
Ta luôn có: \(-\left(\sqrt{x}-2\right)^2\le0\) với \(x\ge0;x\ne15+4\sqrt{11}\)
\(\Rightarrow-\left(\sqrt{x}-2\right)^2+11\le11\)
\(\Rightarrow C\ge\dfrac{1}{11}\). Dấu "=" xảy ra <=> x=4 (t/m)
Vậy MinC=\(\dfrac{1}{11}\) khi x=4
Bài 2:
ĐKXĐ: \(x\ge1\)
\(\sqrt{36x-36}-\sqrt{9x-9}-\sqrt{4x-4}=16-\sqrt{x-1}\)
\(\Leftrightarrow6\sqrt{x-1}-3\sqrt{x-1}-2\sqrt{x-1}=16-\sqrt{x-1}\)
\(\Leftrightarrow2\sqrt{x-1}=16\)
\(\Leftrightarrow4\left(x-1\right)=256\)
\(\Leftrightarrow x-1=64\)
\(\Leftrightarrow x=65\left(tmđk\right)\)
Vậy phương trình đã cho có nghiệm x=65
bài 3:
ta có:
\(A=\dfrac{a-b}{b^2}\sqrt{\dfrac{a^2b^4}{a^2-2ab+b^2}}\)
\(A=\dfrac{a-b}{b^2}.\dfrac{ab^2}{a-b}=a\)