Bài 1: Rút gọn biểu thức
a)\(\left(4x-1\right)^2+\left(3x+1\right)^2+2.\left(4x-1\right).\left(3x+1\right)\)
b)\(\left(x^2+2\right).\left(x-5\right)+\left(x-5\right).\left(x^2+5x+25\right)\)
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2 tháng 10 2021
b: Ta có: \(\left(4x-y\right)\left(4x+y\right)-2\left(3x-2y\right)^2+\left(x-3y\right)^2\)
\(=16x^2-y^2-2\left(9x^2-12xy+4y^2\right)+x^2-6xy+9y^2\)
\(=17x^2-6xy+8y^2-18x^2+24xy-8y^2\)
\(=-x^2+18xy\)
c: Ta có: \(\left(2a-3b+4c\right)\left(2a-3b-4c\right)\)
\(=\left(2a-3b\right)^2-16c^2\)
\(=4a^2-12ab+9b^2-16c^2\)
25 tháng 2 2023
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
DN
12 tháng 6 2018
\(a,\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)=9x^2+30x+25+9x^2-30x+25-9x^2+4=9x^2+54\)
\(b,BT=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2+25x\)
\(c,BT=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-z-x-y\right)^2=z^2\)
2 tháng 10 2021
a: Ta có: \(\left(3x-1\right)^2-2\left(5x-2\right)^2-2\left(x^2+x-1\right)\left(x-1\right)\)
\(=9x^2-6x+1-2\left(25x^2-20x+4\right)-2\left(x^3-x^2+x^2-x-x+1\right)\)
\(=9x^2-6x+1-50x^2+40x-8-2\left(x^3-2x+1\right)\)
\(=-41x^2+34x-7-2x^3+4x-2\)
\(=-2x^3-41x^2+38x-9\)
b: Ta có: \(\left(3a+1\right)^2+2\left(9a^2-1\right)+\left(3a-1\right)^2\)
\(=\left(3a+1+3a-1\right)^2\)
\(=36a^2\)
22 tháng 12 2023
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
QN
Rút gọn biểu thức:
\(\left(x+3\right)^2+\left(2x+1\right)\left(3x-5\right)-2x\left(3-x\right)+4x+25\)
0
H
12 tháng 6 2018
a)\(9x^2+30x+25+9x^2-30x+25-\left(9x^2-2^2\right)\)
=\(9x^2+54\)=\(9\left(x^2+6\right)\)
b)\(2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)\)
=\(8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\)
=\(x^3-16x^2+25x\)
c)\(\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2\)
=\(\left(x+y-z-\left(x+y\right)\right)^2\)=\(\left(-z\right)^2\)
R
27 tháng 7 2020
a) \(\left(1+x\right)^2+\left(1-x\right)^2\)
\(=1+2x+x^2+1-2x+x^2\)
\(=2x^2+2\)
b) \(\left(x+2\right)^2+\left(1+x\right)\left(1-x\right)\)
\(=x^2+4x+4+1-x^2\)
\(=4x+5\)
c) \(\left(x-3\right)^2+3\left(x+1\right)^2\)
\(=x^2-6x+9+3x^2+6x+3\)
\(=4x^2+12\)
d)\(\left(2+3x\right)\left(3x-2\right)-\left(3x+1\right)^2\)
\(=9x^2-4-9x^2-6x-1\)
\(=-6x-5\)
e) \(\left(x+5\right)\left(x-2\right)-\left(x+2\right)^2\)
\(=x^2-2x+5x-10-x^2-4x-4\)
\(=-x-14\)
f) \(\left(x+3\right)\left(2x-5\right)-2\left(1+x\right)^2\)
\(=2x^2-5x+6x-15-2-4x-2x^2\)
\(=-3x-17\)
g) \(\left(4x-1\right)\left(4x+1\right)-4\left(1-2x\right)^2\)
\(=16x^2-1-4+16x-16x^2\)
\(=16x-5\)
#Học tốt!
Lời giải:
a)
\((4x-1)^2+(3x+1)^2+2(4x-1)(3x+1)\)
\(=(4x-1)^2+2(4x-1)(3x+1)+(3x+1)^2\)
\(=[(4x-1)+(3x+1)]^2=(7x)^2=49x^2\)
b)
\((x^2+2)(x-5)+(x-5)(x^2+5x+25)\)
\(=(x-5)[(x^2+2)+(x^2+5x+25)]\)
\(=(x-5)(2x^2+5x+27)\)