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a) \(A=5x\left(4x^2-2x+1\right)-2x\left(10x^2-5x-2\right)\)
\(A=20x^3-10x^2+5x-20x^3+10x^2+4x\)
\(A=9x\)
Thay x = 15 vào, ta có:
\(A=9.15=135\)
b) \(B=5x\left(x-4y\right)-4y\left(y-5x\right)\)
\(B=5x^2-20xy-4y^2+20xy\)
\(B=5x^2-4y\)
Thay \(x=-\frac{1}{5};y=-\frac{1}{2}\) vào, ta có:
\(B=5.\left(-\frac{1}{5}\right)^2-4.\left(-\frac{1}{2}\right)=\frac{11}{5}\)
c) \(C=6xy\left(xy-y^2\right)-8x^2\left(x-y^2\right)-5y^2\left(x^2-xy\right)\)
\(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)
\(C=9x^2y^2-xy^3-8x^3\)
Thay \(x=\frac{1}{2};y=2\) vào, ta có:
\(C=9.\left(\frac{1}{2}\right)^2.2^2-\frac{1}{2}.2^3-8.\left(\frac{1}{2}\right)^3=4\)
d) \(D=\left(3x+5\right)\left(2x-1\right)+\left(4x-1\right)\left(3x+2\right)\)
\(D=6x^2-3x+10x-5+12x^2+8x-3x-2\)
\(D=18x^2+12x-7\)
Ta có: \(\left|2\right|=\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)
+) Với x = -2
\(D=18.\left(-2\right)^2+12.\left(-2\right)-7=41\)
+) Với x = 2
\(D=18.2^2+12.2-7=89\)
A = (x - 1)(x + 3) - (x - 2)(5x - 4)
A = x2 + 2x - 3 - 5x2 + 14x - 8
A = -4x2 + 16x - 11
B = (3a - 2b)(9a2 + 6ab - 4b2)
B = 27a3 + 18a2b - 12ab2 - 18a2b - 12ab2 + 8b3
B = 27a3 -24ab2 + 8b3
C = (x - 1)(x + 1) - (2x - 3)(4 - 5x)
C = x2 - 1 - 8x + 10x + 12 - 15x
C = x2 - 13x + 11
a/ \(A=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\)
Thay x = 15 vào bt A ta có
A = 9 . 15 = 135
b/ \(B=5x^2-20xy-4y^2+2xy=5x^2-4y^2\)
Thay x = -1/5 ; y = - 1/2 vào bt B ta có
\(B=5.\dfrac{1}{25}-4.\dfrac{1}{4}=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
c/ \(C=6x^2y^2-6xy^3-8x^3+8x^2y^2-5x^2y^2+5xy^3\)
\(=9x^2y^2-xy^3-8x^3\)
Thay x = 1/2 ; y = 2 vào bt C ta có
\(C=9.4.\dfrac{1}{4}-\dfrac{1}{2}.8-8.\dfrac{1}{8}=9-4-1=4\)
d/ \(D=6x^2+10x-3x-5+6x^2-3x+8x-2\)
\(=12x^2+12x-3\)
\(\left|x\right|=2\Rightarrow x=\pm2\)
Thay x = 2 vào bt D có
\(D=12.4+12.2-3=69\)
Thay x = - 2 vào bt D ta có
\(D=12.4-12.2-3=21\)
Bài 1:
\(P=2a^2-2b^2-a^2+2ab-b^2+a^2+2ab+b^2+b^2=2a^2-b^2+4ab\\ Q=\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(2x-3\right)\left(2x+3\right)\\ Q=\left(2x+3-2x+3\right)^2=9^2=81\)
Bài 2:
\(Sửa:A=x^2+2xy+y^2-4x-4y+2=\left(x+y\right)^2-4\left(x+y\right)+4-2\\ A=\left(x+y-2\right)^2-2=\left(3-2\right)^2-2=1-2=-1\)
a: Ta có: \(\left(x+3\right)\left(x+4\right)\left(x+5\right)\left(x+6\right)+1\)
\(=\left(x^2+9x+18\right)\left(x^2+9x+20\right)+1\)
\(=\left(x^2+9x\right)^2+38\left(x^2+9x\right)+360+1\)
\(=\left(x^2+9x\right)^2+2\cdot\left(x^2+9x\right)\cdot19+19^2\)
\(=\left(x^2+9x+19\right)^2\)
b. \(x^2+y^2+2x+2y+2\left(x+1\right)\left(y+1\right)+2\)
\(=\left(x^2+2x+1\right)+2\left(x+1\right)\left(y+1\right)+\left(y^2+2y+1\right)\)
\(=\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\)
\(=\left(x+1+y+1\right)^2=\left(x+y+2\right)^2\)
c. \(x^2-2x\left(y+2\right)+y^2+4y+4\)
\(=x^2-2x\left(y+2\right)+\left(y+2\right)^2\)
\(=\left(x-y-2\right)^2\)
d. \(x^2+2x\left(y+1\right)+y^2+2y+1\)
\(=x^2+2x\left(y+1\right)+\left(y+1\right)^2\)
\(=\left(x+y+1\right)^2\)
\(\dfrac{1}{2}\left(6x-2y\right)\left(3x+y\right)=\dfrac{1}{2}.2\left(3x-y\right)\left(3x+y\right)=9x^2-y^2\)
\(\left(\dfrac{2}{3}z-\dfrac{2}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}x\right).\dfrac{1}{2}=\left(\dfrac{1}{3}z-\dfrac{1}{5}x\right)\left(\dfrac{1}{3}z+\dfrac{1}{5}z\right).2.\dfrac{1}{2}=\dfrac{1}{9}z^2-\dfrac{1}{25}x^2\)
\(\left(5y-3x\right).\dfrac{1}{4}\left(12x+20y\right)=\left(5y-3x\right)\left(5y+3x\right).4.\dfrac{1}{4}=25y^2-9x^2\)
\(\left(\dfrac{3}{4}y-\dfrac{1}{2}x\right)\left(x+\dfrac{3}{2}y\right)=\left(\dfrac{3}{2}y-x\right)\left(\dfrac{3}{2}y+x\right)=\dfrac{9}{4}y^2-x^2\)
\(\left(a+b+c\right)\left(a+b+c\right)=\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
\(\left(x-y+z\right)\left(x+y-z\right)=x^2-\left(y-z\right)^2=x^2-y^2-z^2+2yz\)
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rồi tìm x
nha bn
a: Sửa đề: \(A=\left(3a-1\right)\left(9a^2+3a+1\right)-\left(3a+1\right)\left(9a^2-3a+1\right)+2a+2\)
\(=27a^3-1-27a^3-1+2a+2=2a=2\cdot5=10\)
b: \(=4x^2+2x+1-20x^3+10x^2+4x\)
\(=-20x^3+14x^2+6x+1\)
c: \(=5x^2-20xy-4y^2+20xy=5x^2-4y^2\)
\(=5\cdot\dfrac{1}{25}-4\cdot\dfrac{1}{4}=\dfrac{1}{5}-1=-\dfrac{4}{5}\)