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a). \(\sqrt{9\left(x-1\right)^2}-12=0\)
\(\Leftrightarrow\sqrt{9}.\sqrt{\left(x-1\right)^2}-12=0\)
\(\Leftrightarrow3x-3-12=0\Leftrightarrow3x=15\Leftrightarrow x=5\)
b). \(\sqrt{2-3x}=10\Leftrightarrow\sqrt{\left(2-3x\right)^2}=10^2\)
\(\Leftrightarrow2-3x=100\Leftrightarrow-3x=18\Leftrightarrow x=-6\)
c). \(\sqrt{x^2-4}=\sqrt{2-x}\Leftrightarrow\left(\sqrt{x^2}-4\right)^2=\left(\sqrt{2-x}\right)^2\)
\(\Leftrightarrow x^2-4=2-x\Leftrightarrow x^2+x-6=0\)
Vậy nghiệm của PT là \(x_1=2,x_2=-3\)
a) \(\sqrt{9\left(x-1\right)^2}-12=0\Leftrightarrow3\left|x-1\right|-12=0\)(1)
Nếu x\(\ge1\) thì
(1)\(\Leftrightarrow3\left(x-1\right)-12=0\Leftrightarrow3x-3-12=0\Leftrightarrow3x-15=0\Leftrightarrow3x=15\Leftrightarrow x=5\left(tm\right)\)
Nếu x<1 thì \(\left(1\right)\Leftrightarrow3\left(1-x\right)-12=0\Leftrightarrow3-3x-12=0\Leftrightarrow-9=3x\Leftrightarrow x=-3\left(tm\right)\)Vậy S={-3;5)
b) ĐK: x\(\le\dfrac{2}{3}\)
\(\sqrt{2-3x}=10\Leftrightarrow2-3x=10^2\Leftrightarrow2-3x=100\Leftrightarrow3x=-98\Leftrightarrow x=\dfrac{-98}{3}\left(tm\right)\)
Vậy S={\(-\dfrac{98}{3}\)}
c) Ta có ĐK của \(\sqrt{x^2-4}\) là \(x\ge2\) hoặc \(x\le-2\)
ĐK của \(\sqrt{2-x}\) là \(x\le2\)
Vậy muốn xảy ra dấu '=' thì ĐK là \(x\le-2\) hoặc x=2
Ta thử và thấy x=2 là nghiệm cảu phương trình
Ta có \(\sqrt{x^2-4}=\sqrt{2-x}\Leftrightarrow x^2-4=2-x\Leftrightarrow x^2+x-6=0\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-3\left(tm\right)\end{matrix}\right.\)
Vậy S={-3;2}
Bài 1:
a: Ta có: \(\sqrt{3x^2}=\sqrt{12}\)
\(\Leftrightarrow3x^2=12\)
\(\Leftrightarrow x^2=4\)
hay \(x\in\left\{2;-2\right\}\)
b: Ta có: \(\sqrt{\left(x-2\right)^2}=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
Sửa lại câu c) đặt \(\sqrt{x}+1=\)t \(\Rightarrow\left[2\left(t+\dfrac{1}{2}\right)\right]\left(t-3\right)\)=7⇒\(\left\{{}\begin{matrix}t=3\\t=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\\x=\dfrac{9}{4}\end{matrix}\right.\)
a) \(\left(\sqrt{4-3x}\right)^2=8^2\)\(\Leftrightarrow4-3x=64\Rightarrow x=-20\)
b) \(\sqrt{4x-8}+1=12\sqrt{\dfrac{x-2}{9}}\Leftrightarrow2\sqrt{x-2}+1\)\(=\left(12\sqrt{\left(x-2\right).\dfrac{1}{9}}\right)\)
\(\Leftrightarrow2t+1=12.\dfrac{1}{3}t\) (Đặt t = \(\sqrt{x-2}\))
\(\Rightarrow t=\dfrac{1}{2}\) \(\Rightarrow\sqrt{x-2}=\dfrac{1}{2}\)\(\Rightarrow x=\dfrac{9}{4}\)
c) pt\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x}+1=7\\\sqrt{x}-2=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=3\\\sqrt{x}=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=9\\x=4\end{matrix}\right.\)
a: Ta có: \(\sqrt{4-3x}=8\)
\(\Leftrightarrow4-3x=64\)
\(\Leftrightarrow3x=-60\)
hay x=-20
b: ta có: \(\sqrt{4x-8}-12\sqrt{\dfrac{x-2}{9}}=-1\)
\(\Leftrightarrow2\sqrt{x-2}-12\cdot\dfrac{\sqrt{x-2}}{3}=-1\)
\(\Leftrightarrow x-2=\dfrac{1}{4}\)
hay \(x=\dfrac{9}{4}\)
\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right)\sqrt{9-x^2}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\left(x+2\right)\left(x+3\right)+x\sqrt{\left(3-x\right)\left(3+x\right)}}{x\left(3-x\right)+\left(x+2\right)\sqrt{\left(3-x\right)\left(3+x\right)}}\)
\(=\frac{\sqrt{3+x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}{\sqrt{3-x}\left(\left(x+2\right)\sqrt{x+3}+x\sqrt{3-x}\right)}\)
\(=\frac{\sqrt{3+x}}{\sqrt{3-x}}\)
\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right)\sqrt{x^2-6x+8}}\)
\(=\frac{\left(x-3\right)\left(x-2\right)+3\sqrt{\left(x-4\right)\left(x-2\right)}}{3\left(x-4\right)+\left(x-3\right)\sqrt{\left(x-4\right)\left(x-2\right)}}\)
\(=\frac{\sqrt{x-2}\left(\left(x-3\right)\sqrt{x-2}+3\sqrt{x-4}\right)}{\sqrt{x-4}\left(3\sqrt{x-4}+\left(x-3\right)\sqrt{x-2}\right)}\)
\(=\frac{\sqrt{x-2}}{\sqrt{x-4}}\)
\(a)\)\(\sqrt{9\left(x-1\right)^2}-12=0\)
\(\Leftrightarrow\)\(\sqrt{\left[3\left(x-1\right)\right]^2}=12\)
\(\Leftrightarrow\)\(\left|3\left(x-1\right)\right|=12\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3\left(x-1\right)=12\\3\left(x-1\right)=-12\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)
Vậy \(x=-3\) hoặc \(x=5\)
\(b)\) ĐKXĐ : \(x\le\frac{3}{2}\)
\(\sqrt{2-3x}=10\)
\(\Leftrightarrow\)\(2-3x=100\)
\(\Leftrightarrow\)\(3x=-98\)
\(\Leftrightarrow\)\(x=\frac{-98}{3}\) ( không thỏa mãn )
Vậy không có x thỏa mãn đề bài
\(c)\) ĐKXĐ : \(x>2\) hoặc \(x\le-2\)
\(\sqrt{x^2-4}=\sqrt{2-x}\)
\(\Leftrightarrow\)\(\sqrt{\left(x-2\right)\left(x+2\right)}-\sqrt{2-x}=0\)
\(\Leftrightarrow\)\(\sqrt{2-x}.\sqrt{-x-2}-\sqrt{2-x}=0\)
\(\Leftrightarrow\)\(\sqrt{2-x}\left(\sqrt{-x-2}-1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}-\sqrt{2-x}=0\\\sqrt{-x-2}-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(loai\right)\\x=-3\left(nhan\right)\end{cases}}}\)
Vậy \(x=-3\)
Chúc bạn học tốt ~
PS : mới lớp 8 :v làm sai thì thông cảm