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PN
10 tháng 5 2021
b, \(15\left(x+3\right)+20x\left(x+8\right)=15x+45+20x^2+160x\)
\(=20x^2+175x+45=...\)
c, \(6\left(x-9\right)-3x\left(y-x\right)=6x-54-3xy+3x^2\)
d, \(2xy+10x^2-x\) không phân tích được nữa nhé
e, \(4ab^2-28a+16b\)không phân tích được nữa nhé
g, \(a\left(a+b\right)=ab\left(a+b\right)< =>\left(a+b\right)\left(a-ab\right)=0< =>\left(a+b\right)a\left(1-b\right)=0\)
h, \(30a^2+6a-6=\left(\sqrt{30}a\right)^2+2.\sqrt{30}.\frac{3}{\sqrt{30}}+\frac{3}{10}-\frac{63}{10}\)
\(=\left(\sqrt{30}a+\frac{3}{\sqrt{30}}\right)^2-\sqrt{\frac{63}{10}}^2=\left(\sqrt{30}a+\frac{3}{\sqrt{30}}-\sqrt{\frac{63}{10}}\right)\left(\sqrt{30}a+\frac{3}{\sqrt{30}}+\sqrt{\frac{63}{10}}\right)\)
ND
0
9 tháng 12 2019
Ta có:
\(\frac{x}{x^2+x+1}=-\frac{1}{4}\Rightarrow x^2+x+1=-4x\)
\(\Rightarrow x^2+5x+1=0\Rightarrow x^2=5x+1\)
Với x2=5x+1 ta được:
\(P=\frac{2x\left(5x+1\right)^2+10\left(5x+1\right)^2+2x\left(5x+1\right)-7\left(5x+1\right)-35x+2009}{2029+60x+11\left(5x+1\right)-5x\left(5x+1\right)-\left(5x+1\right)^2}\)
\(P=\frac{2x\left(25x^2+10x+1\right)+10\left(25x^2+10x+1\right)+10x^2+2x-35x-7-35x+2009}{2029+60x+55x+11-25x^2-5x-\left(25x^2+10x+1\right)}\)
\(P=\frac{50x^3+20x^2+2x+250x^2+100x+10+10x^2+2x-35x-7-35x+2009}{2029+60x+55x+11-25x^2-5x-25x^2-10x-1}\)
\(P=\frac{50x^3+280x^2+34x+2012}{2039+100x-50x^2}\)
\(P=\frac{50x\left(5x+1\right)+280\left(5x+1\right)+34x+2012}{2039+100x-50\left(5x+1\right)}\)
\(P=\frac{250x^2+50x+1400x+280+34x+2012}{2039+100x-250x-50}\)
\(P=\frac{250\left(5x+1\right)+50x+1400x+280+34x+2012}{1989-150x}\)
\(P=\frac{1250x+250+50x+1400x+280+34x+2012}{1989-150x}\)
CT
0
NT
2
15 tháng 4 2018
Ta có : \(P=\frac{x^2-10x+22}{\left(x-3\right)^2}\)
Đặt : \(x-3=y\Leftrightarrow x=y+3\)
\(P=\frac{\left(y+3\right)^2-10\left(y+3\right)+22}{y^2}\)
\(P=\frac{y^2+6y+9-10y-30+22}{y^2}\)
\(P=\frac{y^2-4y+1}{y^2}\)
\(P=\frac{y^2}{y^2}-\frac{4y}{y^2}+\frac{1}{y^2}\)
\(P=1-\frac{4}{y}+\frac{1}{y^2}\)
\(P=\left(\frac{1}{y^2}-\frac{4}{y}+4\right)-3\)
\(P=\left(\frac{1}{y}-2\right)^2-3\)
Mà \(\left(\frac{1}{y}-2\right)^2\ge0\forall y\)
\(\Rightarrow P\ge-3\)
Dấu "=" xảy ra khi :
\(\frac{1}{y}-2=0\Leftrightarrow\frac{1}{y}=2\Leftrightarrow y=\frac{1}{2}\)
Lại có : \(x=y+3\)
\(\Rightarrow x=\frac{7}{2}\)
Vậy \(P_{Min}=-3\Leftrightarrow x=\frac{7}{2}\)
13 tháng 6 2019
a) \(3y^2\left(2y-1\right)+y-y\left(1-y+y^2\right)-y^2+y \)
= \(6y^3-3y^2+y-y+y^2-y^3-y^2+y\)
= \(5y^3-3y^2+y\)
b)\(25x-4\left(3x-1\right)+\left(5-2x\right)7\)
= \(25x-12x+4+35-14x\)
= \(-x+39\)
c) \(11x-2\left(10x-1\right)-\left(4x-1\right)\left(-2\right)\)
= \(11x-\left(20x-2\right)-\left(-8x+2\right)\)
= \(11x-20x+2+8x-2\)
= \(-x\)
d) \(\left(\frac{1}{2x}\right)3-x\left(1-2x-\frac{1}{8x^2}\right)-x\left(x+\frac{1}{2}\right)\)
= \(\frac{3}{2x}-x+2x^2+\frac{x}{8x^2}-x^2-\frac{x}{2}\)
= \(\left(\frac{3}{2x}+\frac{1}{8x}-\frac{x}{2}\right)+x^2-x\)
= \(\left(\frac{12+1-4x^2}{8x}\right)+x^2-x\)
= \(\frac{13-4x^2}{8x}+\frac{8x^3}{8x}-\frac{8x^2}{8x}\)
= \(\frac{13-4x^2+8x^3-8x^2}{8x}\)
= \(\frac{8x^3-12x^2+13}{8x}\)
= x2 - \(\frac{3}{2}\)+\(\frac{13}{8x}\)
e) \(12\left(2-3x\right)+35x-\left(x+1\right)\left(-5\right)\)
= \(24-36x+35x-\left(-5x-5\right)\)
= \(24-36x+35x+5x+5\)
= 4x + 29
11 tháng 8 2020
a) Ta có: \(\left(4x^2-12x+9\right)-1\)
\(=\left(2x-3\right)^2-1^2\)
\(=\left(2x-3-1\right)\left(2x-3+1\right)\)
\(=\left(2x-4\right)\left(2x-2\right)\)
\(=4\left(x-2\right)\left(x-1\right)\)
b) Ta có: \(\left(\frac{x^2}{4}+2xy+4y^2\right)-25\)
\(=\left[\left(\frac{x}{2}\right)^2+2\cdot\frac{x}{2}\cdot2y+\left(2y\right)^2\right]-5^2\)
\(=\left(\frac{x}{2}+2y\right)^2-5^2\)
\(=\left(\frac{x}{2}+2y-5\right)\left(\frac{x}{2}+2y+5\right)\)
c) Ta có: \(1+12x+35x^2\)
\(=35x^2+12x+1\)
\(=35x^2+5x+7x+1\)
\(=5x\left(7x+1\right)+\left(7x+1\right)\)
\(=\left(7x+1\right)\left(5x+1\right)\)
d) Ta có: \(9x^2-24xy+15y^2\)
\(=9x^2-9xy-15xy+15y^2\)
\(=9x\left(x-y\right)-15y\left(x-y\right)\)
\(=\left(x-y\right)\left(9x-15y\right)\)
\(=3\left(x-y\right)\left(3x-5y\right)\)
e) Ta có: \(25x^2-20xy+3y^2\)
\(=25x^2-15xy-5xy+3y^2\)
\(=5x\left(5x-3y\right)-y\left(5x-3y\right)\)
\(=\left(5x-3y\right)\left(5x-y\right)\)
f) Ta có: \(24x^4-10x^2y+y^2\)
\(=24x^4-4x^2y-6x^2y+y^2\)
\(=4x^2\left(6x^2-y\right)-y\left(6x^2-y\right)\)
\(=\left(6x^2-y\right)\left(4x^2-y\right)\)
\(35x^2-49=10x^2< =>25x^2-7^2=0\)
\(< =>\left(5x\right)^2-7^2=0< =>\left(5x+7\right)\left(5x-7\right)=0\)
\(=>\left[{}\begin{matrix}x=-\dfrac{7}{5}\\x=\dfrac{7}{5}\end{matrix}\right.\)
Ta có: \(35x^2-49=10x^2\)
\(\Leftrightarrow25x^2-49=0\)
\(\Leftrightarrow\left(5x-7\right)\left(5x+7\right)=0\)
\(\Leftrightarrow x\in\left\{\dfrac{7}{5};-\dfrac{7}{5}\right\}\)