g) x^2+y^2+2(x+y)+2=0
h) 4x+y^2-4x-4y+6=0
i) x^2-7x+12=0
k) 1/2 × x+7/8x=11
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Những câu hỏi liên quan
22 tháng 10 2018
a/ \(3x+3y-4x-4y=3\left(x+y\right)-4\left(x+y\right)=\left(x+y\right)\left(3-4\right)=-1\left(x+y\right)\)
b/ \(7x\left(x-y\right)-\left(y-x\right)=7x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(7x+1\right)\)
c/ \(5x\left(1-x\right)+\left(x-1\right)=5x\left(1-x\right)-\left(1-x\right)=\left(1-x\right)\left(5x-1\right)\)
d/ \(4x\left(x-y\right)+3\left(x-y\right)^2=\left(x-y\right)\left(4x+3x-3y\right)=\left(x-y\right)\left(7x-3y\right)\)
e/ \(4x\left(x-y\right)+3\left(y-x\right)^2=4x\left(x-y\right)+3\left(x-y\right)^2=\left(x-y\right)\left(4x+3x-3y\right)=\left(x-y\right)\left(7x-3y\right)\)
g/ \(x^2+8x+7=x^2+x+7x+7=x\left(x+1\right)+7\left(x+1\right)=\left(x+1\right)\left(x+7\right)\)
h/ \(x^2-6x-16=x^2+2x-8x-16=x\left(x+2\right)-8\left(x+2\right)=\left(x+2\right)\left(x-8\right)\)
i/ \(4x^2-8x+3=4x^2-2x-6x+3=2x\left(2x-1\right)-3\left(2x-1\right)=\left(2x-1\right)\left(2x-3\right)\)
k/ \(3x^2-11x+6=3x^2-9x-2x+6=3x\left(x-3\right)-2\left(x-3\right)=\left(x-3\right)\left(3x-2\right)\)
15 tháng 10 2021
a: \(x^2\left(x-3\right)-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
b: \(2a\left(x+y\right)+x+y=\left(x+y\right)\left(2a+1\right)\)
c: \(6x^2-12x-7x+14\)
\(=6x\left(x-2\right)-7\left(x-2\right)\)
\(=\left(x-2\right)\left(6x-7\right)\)
TG
19 tháng 10 2020
a) \(x^2-8x+20\)
\(=x^2-2.x.4+16+4\)
\(=\left(x-4\right)^2+4\)
Có: \(\left(x-4\right)^2\ge0\Rightarrow\left(x-4\right)^2+4>0\)
Hay:.............
b) \(x^2+11\)
Có: \(x^2\ge0\Rightarrow x^2+11>0\)
Hay:.............
c) \(4x^2-12x+11\)
\(=4\left(x^2-3x+\frac{11}{4}\right)\)
\(=4\left(x^2-2.x.\frac{3}{2}+\frac{9}{4}+\frac{1}{2}\right)\)
\(=4\left(x-\frac{3}{2}\right)^2+2>0\)
d) \(x^2+5y^2+2x+6y+34\)
\(=x^2+2.x.1+1+y^2+4y^2+2.y.3+9+24\)
\(=\left(x^2+2.x.1+1\right)+\left(y^2+2.y.3+9\right)+4y^2+24\)
\(=\left(x+1\right)^2+\left(y+3\right)^2+\left(2y\right)^2+24\)
Ta có: \(\left\{{}\begin{matrix}\left(x+1\right)^2\ge0\\\left(y+3\right)^2\ge0\\\left(2y\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x+1\right)^2+\left(y+3\right)^2+\left(2y\right)^2+24>0\)
f) \(x^2-2x+y^2+4y+6\)
\(=x^2-2.x.1+1+y^2+2.y.2+4+1\)
\(=\left(x-1\right)^2+\left(y+2\right)^2+1>0\)
PT
29 tháng 8 2018
Bài 2;
\(a)x^4-16x=0\Rightarrow x^4=16x\Leftrightarrow x^3=16\Leftrightarrow x=\sqrt[3]{16}\)
PT
29 tháng 8 2018
\(c)4x^2-\frac{1}{4}=0\Leftrightarrow4x^2=\frac{1}{4}\Leftrightarrow x^2=\frac{1}{16}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\x=-\frac{1}{4}\end{cases}}\)