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2.
\(y=3x-1\) // \(y=3x-3\) vì \(\left\{{}\begin{matrix}3=3\\-1\ne-3\end{matrix}\right.\)
3.
\(a,y=\left(m+2\right)x-3//y=2x\Leftrightarrow\left\{{}\begin{matrix}m+2=2\\-3\ne0\end{matrix}\right.\Leftrightarrow m=0\)
\(b,\) Thay \(x=1;y=2\Leftrightarrow m+2-3=2\Leftrightarrow m=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a: \(\Leftrightarrow\left\{{}\begin{matrix}2x-4y=10\\3x+4y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-1\end{matrix}\right.\)
b: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{3}x+3y=\sqrt{3}\\2\sqrt{3}x+2y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\sqrt{3}\\x=1\end{matrix}\right.\)
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![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\)
\(\Leftrightarrow A^3=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\cdot\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(\Leftrightarrow A^3=4+3\cdot\left(-1\right)\cdot A\)
\(\Leftrightarrow A^3=4-3A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A^2+A^2-A+4A-4=0\)
\(\Leftrightarrow A^2\left(A-1\right)+A\left(A-1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\) \(\left(d\right)\) đi qua \(A\left(1;2\right)\Leftrightarrow x=1;y=2\)
\(\Leftrightarrow2=m+1-2m+3\Leftrightarrow m=2\)
\(b,m=2\Leftrightarrow\left(d\right):y=3x-2\cdot2+3=3x-1\)
\(y=2\Leftrightarrow x=1\Leftrightarrow A\left(1;2\right)\\ y=5\Leftrightarrow x=2\Leftrightarrow B\left(2;5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bạn ơi giúp mình vs
tìm x,y, z nguyên thỏa mãn
x^3 + xyz = 957
y^3 + xyz = 759
z^3 + xyz = 579
\(\Leftrightarrow x-3\sqrt{x}-\sqrt{x-8}+1=0\)
\(\Leftrightarrow x=9\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: Thay x=16 vào A, ta được:
\(A=\dfrac{16+3}{4-2}=\dfrac{19}{2}\)
Câu 2:
a) ĐKXĐ: \(x\ge0\)
\(pt\Leftrightarrow\sqrt{3x}-10\sqrt{3x}+21\sqrt{3x}=12\)
\(\Leftrightarrow12\sqrt{3x}=12\Leftrightarrow\sqrt{3x}=1\Leftrightarrow3x=1\Leftrightarrow x=\dfrac{1}{3}\left(tm\right)\)
b) ĐKXĐ: \(x\ge2\)
\(pt\Leftrightarrow2\sqrt{x-2}-4\sqrt{x-2}=-1\)
\(\Leftrightarrow-2\sqrt{x-2}=-1\Leftrightarrow\sqrt{x-2}=\dfrac{1}{2}\)
\(\Leftrightarrow x-2=\dfrac{1}{4}\Leftrightarrow x=\dfrac{9}{4}\)
Câu 3:
\(A=\dfrac{5\sqrt{x}-25x}{1-5\sqrt{x}}=\dfrac{5\sqrt{x}\left(1-5\sqrt{x}\right)}{1-5\sqrt{x}}=5\sqrt{x}\)