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PN
1
26 tháng 9 2021
a: Ta có: \(\left(x-2\right)^3-x\left(x+1\right)\left(x-1\right)+6x^2=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2=5\)
\(\Leftrightarrow13x=13\)
hay x=1
2 tháng 12 2023
bài 5:
1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)
2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)
\(=\dfrac{5x\left(2y-x\right)}{-2\left(x+2y\right)\left(2y-x\right)}=\dfrac{-5x}{2\left(x+2y\right)}\)
3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\dfrac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)
\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
4: \(\dfrac{\left(x+1\right)\left(x^2-2x+1\right)}{\left(6x^2-6\right)\left(x^3-1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)^2}{6\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{6\left(x-1\right)\left(x+1\right)\cdot\left(x^2+x+1\right)}\)
\(=\dfrac{1}{6\left(x^2+x+1\right)}\)
5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)
\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)
\(=-\dfrac{2x^2-6x-x+3}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{2x\left(x-3\right)-\left(x-3\right)}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{\left(x-3\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{-x+3}{2x+1}\)
Bài 3:
1: \(9x^3-xy^2\)
\(=x\cdot9x^2-x\cdot y^2\)
\(=x\left(9x^2-y^2\right)\)
\(=x\left(3x-y\right)\left(3x+y\right)\)
2: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
3: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
4: \(6xy-x^2+36-9y^2\)
\(=36-\left(x^2-6xy+9y^2\right)\)
\(=36-\left(x-3y\right)^2\)
\(=\left(6-x+3y\right)\left(6+x-3y\right)\)
5: \(x^4-6x^2+5\)
\(=x^4-x^2-5x^2+5\)
\(=x^2\left(x^2-1\right)-5\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x-1\right)\left(x+1\right)\)
6: \(9x^2-6x-y^2+2y\)
\(=\left(9x^2-y^2\right)-\left(6x-2y\right)\)
\(=\left(3x-y\right)\left(3x+y\right)-2\left(3x-y\right)\)
\(=\left(3x-y\right)\left(3x+y-2\right)\)
NT
0
TA
18 tháng 12 2020
\((x-2)^3-(x+5)(x^2-5x+25)+6x^2=11\\\Leftrightarrow (x-2)^3-(x+5)(x^2-5.x+5^2)+6x^2=11 \\\Leftrightarrow x^3-6x^2+12x-8 -(x^3+5^3)+6x^2-11=0 \\\Leftrightarrow 12x-144=0 \\\Leftrightarrow x=12\)
Vậy \(x=12\).
TG
15 tháng 3 2022
(x−2)3−(x+5)(x2−5x+25)+6x2=11
=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11
=>x3−6x2+12x−8−(x3+53)+6x2−11=0
=>12x−144=0
=>x=12(x−2)3−(x+5)(x2−5x+25)+6x2=11
=>(x−2)3−(x+5)(x2−5.x+52)+6x2=11
=>x3−6x2+12x−8−(x3+53)+6x2−11=0
=>12x−144=0
=>x=12
Vậy x=12x=12.
cho tôi đúng đi
23 tháng 7 2023
a: =>x^2-6x+9+y^2+8y+16=0
=>(x-3)^2+(y+4)^2=0
=>x=3 và y=-4
MH
4 tháng 10 2021
1) \(x^2+6x+9\)
\(=\left(x+3\right)^2\)
2) \(10x-25-x^2\)
\(=-25+10x-x^2\)
\(=-\left(5-x\right)^2\)
3) \(8x^3-\dfrac{1}{8}\)
\(=\left(2x\right)^3-\left(\dfrac{1}{2}\right)^3\)
\(=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
4) \(\dfrac{1}{25}x^2-64y^2\)
\(=\left(\dfrac{1}{5}x\right)^2-\left(8y\right)^2\)
\(=\left(\dfrac{1}{5}x+8y\right)\left(\dfrac{1}{5}x-8y\right)\)
4 tháng 10 2021
\(x^2+6x+9=\left(x+3\right)^2\)
\(10x-25-x^2=-\left(x-5\right)^2\)
\(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
NL
1
15 tháng 12 2023
\(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)
=>\(x^3-6x^2+12x-8-\left(x^3+125\right)+6x^2=11\)
=>\(x^3+12x-8-x^3-125=11\)
=>12x-133=11
=>12x=144
=>\(x=\dfrac{144}{12}=12\)
18 tháng 8 2021
a: Ta có: \(\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(4x^2-9\right)\)
\(=4x^2+12x+9+4x^2-12x+9-8x^2+18\)
\(=36\)
Bài 2:
a: \(\left(y^2+6x^2\right)\left(y^2-6x^2\right)=y^4-36x^4\)
b: \(\left(4x+5\right)\left(16x^2-20x+25\right)=\left(16x^2-25\right)\left(4x-5\right)\)
\(=64x^3-16x^2-100x+125\)
\(x^2-25=6x-9\)
\(\Rightarrow x^2-25-\left(6x-9\right)=0\)
\(\Rightarrow x^2-25-6x+9=0\)
\(\Rightarrow\left(x^2-6x+9\right)-25=0\)
\(\Rightarrow\left(x^2-2x.3+3^2\right)-5^2=0\)
\(\Rightarrow\left(x-3\right)^2-5^2=0\)
\(\Rightarrow\left(x-3-5\right)\left(x-3+5\right)=0\)
\(\Rightarrow\left(x-8\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-8=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=8\\x=-2\end{cases}}}\)