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Bài 1 :
a) \(3x\left(5x^2-2x-1\right)=3x\cdot5x^2+3x\left(-2x\right)+3x\left(-1\right)\)
\(=15x^3-6x^2-3x\)
b) \(\left(x^2-2xy+3\right)\left(-xy\right)\)
\(=x^2\left(-xy\right)-2xy\left(-xy\right)+3\left(-xy\right)\)
\(=-x^3y+2x^2y^2-3xy\)
c) \(\frac{1}{2}x^2y\left(2x^3-\frac{2}{5}xy-1\right)\)
\(=\frac{1}{2}x^2y\cdot2x^3+\frac{1}{2}x^2y\cdot\left(-\frac{2}{5}xy\right)+\frac{1}{2}x^2y\left(-1\right)\)
\(=x^5y-\frac{1}{5}x^3y^2-\frac{1}{2}x^2y\)
d) \(\frac{1}{2}xy\left(\frac{2}{3}x^2-\frac{3}{4}xy+\frac{4}{5}y^2\right)\)
\(=\frac{1}{2}xy\cdot\frac{2}{3}x^2+\frac{1}{2}xy\cdot\left(-\frac{3}{4}xy\right)+\frac{1}{2}xy\cdot\frac{4}{5}y^2\)
\(=\frac{1}{3}x^3y-\frac{3}{8}x^2y^2+\frac{2}{5}xy^3\)
e) \(\left(x^2y-xy+xy^2+y^3\right)\left(3xy^3\right)\)
= \(x^2y\cdot3xy^3-xy\cdot3xy^3+xy^2\cdot3xy^3+y^3\cdot3xy^3\)
\(=3x^3y^4-3x^2y^4+3x^2y^5+3xy^6\)
Bài 2 :
3(2x - 1) + 3(5 - x) = 6x - 3 + 15 - x = (6x - x) - 3 + 15 = 5x - 3 + 15
Thay x = -3/2 vào biểu thức trên ta có : \(5\cdot\left(-\frac{3}{2}\right)-3+15\)
\(=-\frac{15}{2}-3+15=\frac{9}{2}\)
b) 25x - 4(3x - 1) + 7(5 - 2x)
= 25x - 12x + 4 + 35 - 14x
= (25x - 12x - 14x) + 4 + 35 = -x + 4 + 35 = -x + 39
Thay \(x=2\)vào biểu thức trên ta có : -2 + 39 = 37
c) 4x - 2(10x + 1) + 8(x - 2)
= 4x - 20x - 2 + 8x - 16
= (4x - 20x + 8x) - 2 - 16 = -8x - 2 - 16 = -8x - 18
Thay x = 1/2 vào biểu thức trên ta có \(-8\cdot\frac{1}{2}-18=-4-18=-22\)
d) Tương tự
Bài 3:
a) \(2x\left(x-4\right)-x\left(2x+3\right)=4\)
=> 2x2 - 8x - 2x2 - 3x = 4
=> (2x2 - 2x2) + (-8x - 3x) = 4
=> -11x = 4
=> x = \(-\frac{4}{11}\)
b) x(5 - 2x) + 2x(x - 7) = 18
=> 5x - 2x2 + 2x2 - 14x = 18
=> 5x - 14x = 18
=> -9x = 18
=> x = -2
Còn 2 câu làm tương tự
a) 5x2 ( 3x2 -7x+2)-15x(x-3)
=15x4-35x3+10x2-15x2+45x
=15x4-35x3-5x2+45x
c) (x+3)(x-3)(x-2)(x+1)
=(x2-9)(x2+x-2x-2)
=(x2-9)(x2-x-2)
=x4-x3-2x2-9x2+9x+18
=x4-x3-11x2+9x+18
d)(2x+1)2+(4x-1)2+2(2x+1)(4x+1)
=2x2+4x+1-16x2-8x+1
=2x2+4x+1-16x2-8x+1+16x2-4x+8x-2
=2x2+7
e) (2x2-3x)(5x2-2x+1)-10x2(x+3)
=10x4 -4x3+2x2-15x3+6x2-3 -10x2-30x
=10x4-19x3-2x2-30x-3
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/ \(A=20x^3-10x^2+5x-20x^3+10x^2+4x=9x\)
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
\(D=12.4-12.2-3=21\)
a: \(P=\left(5x-1-5x-4\right)^2=\left(-3\right)^2=9\)
b: \(Q=\left(x+y\right)^3-3xy\left(x+y\right)=x^3+y^3\)
c: \(=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(=\dfrac{5^{32}-1}{2}\)
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:
\(Sửa:A=x^2+2xy+y^2-4x-4y+2=\left(x+y\right)^2-4\left(x+y\right)+4-2\\ A=\left(x+y-2\right)^2-2=\left(3-2\right)^2-2=1-2=-1\)
\(a,=2x^2+3x^2y-2x-2x^2+6x^2y-3x=9x^2y-5x\\ b,=\left(x-1\right)\left(x+2-x-5\right)=-3\left(x-1\right)=3-3x\)