Giải thích các bước giải:
1, `x^2+8x=0`
=> `x(x+8)=0`
=> \(\left[ \begin{array}{l}x=0\\x+8=0\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=0\\x=-8\end{array} \right.\)
2, `x^2-2x√2+2=0`
=>`(x-√2)^2=0`
=> `x-√2=0
=>`x=√2`
3, `3x^2-10x+8=0`
`3x^2-4x-6x+8=0`
`x(3x-4)-2(3x-4)=0`
`(3x-4)(x-2)=0`
=>\(\left[ \begin{array}{l}3x-4=0\\x-2=0\end{array} \right.\)
=>\(\left[ \begin{array}{l}x=4/3\\x=2\end{array} \right.\)
4, `-x^4+(√3-√2)x^2=0`
`-x^2+√3x^2-√2x^2=0`
`-x^2(x^2-√3+√2)=0`
=>\(\left[ \begin{array}{l}x^2=0\\x^2- \sqrt[]{3}+\sqrt[]{2}=0 \end{array} \right.\)
=> \(\left[ \begin{array}{l}x=0\\x^2- \sqrt[]{x=±\sqrt[]{\sqrt[]{3}-\sqrt[]{2} } } \end{array} \right.\)
5, $\left \{ {{3x+2y=1} \atop {x+y=1}} \right.$
$\left \{ {{3x+2y=1} \atop {x=1-y}} \right.$
`3(1-y)+2y=1`
`y=2`
=>`x=1-2`
x=-1
6, `(1)/(x-2) + 1 = (5-x)/(x-2)`
=> `x`$\neq$ `2`
`(1)/(x-2) - (5-x)/(x-2) = -1`
`[1- (5-x)]/(x-2) = -1`
`(-4+x)/(x-2) = -1`
`-4+x=-x+2`
`2x=6`
`x=3`
7, Em không chép lại đề tại dài quá ạ
Đặt `1/x+2y=t` , `1/y+2x=u`
=> $\left \{ {{2t+u=3} \atop {4t-3u=1}} \right.$
=> $\left \{ {{-4t-2u=-6} \atop {4t-3u=1}} \right.$
=> `-5u=-5`
=>`u=1`
=>` 2t+1=3` => `t=1`
=> `(t,u)=(1,1)`
=>$\left \{ {{1/x+2y=1} \atop {1/y+2x=1}} \right.$
=> `(x,y) = (1/3,1/3)`
8, $\left \{ {{x^2+y^2=13} \atop {3x^2-2y^2=-6}} \right.$
=> $\left \{ {{x^2=13-y^2} \atop {3x^2-2y^2=-6}} \right.$
=> `3(13-y^3)-2y^2=-6`
=> \(\left[ \begin{array}{l}x=-3\\x=3\end{array} \right.\)
=> \(\left[ \begin{array}{l}x^2=13-(-3)^2\\x^2=13-3^2\end{array} \right.\)
=> \(\left[ \begin{array}{l}x=-2\\x=2\\x=-2\\x=2\end{array} \right.\)
=> \(\left[ \begin{array}{l}(x,y)=(-2,-3)\\(x,y)=(2,-3)\\(x,y)=(-2,3)\\(x,y)=(2,-3)\end{array} \right.\)
10, `3x^4+10x^2+3=0`
Ta có vế trái luôn dương => x ∉ R
Cho em xin hay nhất
