Rút gọn
a) (a+b)^3-(a-b)^3
b) 8x^3+12x^2y+6xy2+y^3
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RH
2 tháng 10 2021
2a) pt <=> (x + 6)^2 = 0
<=> x = -6
b) pt <=> (4x - 1)^2 = 0
<=> x = 1/4
c) pt<=> (x + 1)^3 = 0
<=> x = -1
2 tháng 10 2021
Bài 1:
a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)
\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)
\(=32x^2+18y^2\)
b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)
\(=x^3-6x^2+12x-8-x^3-6x^2-12x-8\)
\(=-12x^2-24\)
2 tháng 10 2021
Bài 2:
a: Ta có: \(x^2+12x+36=0\)
\(\Leftrightarrow x+6=0\)
hay x=-6
b: Ta có: \(16x^2-8x+1=0\)
\(\Leftrightarrow4x-1=0\)
hay \(x=\dfrac{1}{4}\)
2 tháng 10 2021
Bài 1:
a: Ta có: \(A=\left(4x+3y\right)^2+\left(4x-3y\right)^2\)
\(=16x^2+24xy+9y^2+16x^2-24xy+9y^2\)
\(=32x^2+18y^2\)
b: Ta có: \(B=\left(x-2\right)^3-\left(x+2\right)^3\)
\(=x^3-6x^2+12x-8-x^3-6x^2-12x-8\)
\(=-12x^2-24\)
c: Ta có: \(C=\left(x+2y\right)^2+2\left(x+2y\right)\left(x-2y\right)+\left(x-2y\right)^2\)
\(=\left(x+2y+x-2y\right)^2\)
\(=4x^2\)
10 tháng 7 2021
a) Ta có: \(A=\left(7-2x\right)\left(7+2x\right)+\left(2x+7\right)^2\)
\(=7-4x^2+4x^2+28x+49\)
\(=28x+56\)
b) Ta có: \(B=\left(4x-5\right)^2-\left(2x-1\right)\left(8x-5\right)\)
\(=16x^2-40x+25-\left(16x^2-10x-8x+5\right)\)
\(=16x^2-40x+25-16x^2+18x-5\)
\(=-22x+20\)
c) Ta có: \(C=\left(5x-3\right)^2-2\left(5x-3\right)\left(5-5x\right)+\left(5x-5\right)^2\)
\(=\left(5x-3\right)^2+2\cdot\left(5x-3\right)\left(5x-5\right)+\left(5x-5\right)^2\)
\(=\left(5x-3+5x-5\right)^2\)
\(=\left(10x-8\right)^2\)
\(=100x^2-160x+64\)
d) Ta có: \(D=\left(2a+3b-c\right)\left(2a-3b+c\right)-\left(4a^2-9b^2-c^2\right)\)
\(=\left[\left(2a+\left(3b-c\right)\right)\left(2a-\left(3b-c\right)\right)\right]-\left(4a^2-9b^2-c^2\right)\)
\(=4a^2-\left(3b-c\right)^2-4a^2+9b^2+c^2\)
\(=-9b^2+6bc-c^2+9b^2+c^2\)
=6bc
T
24 tháng 10 2023
Bài 1.
a)
\((x-2)(2x-1)-(2x-3)(x-1)-2\\=2x^2-x-4x+2-(2x^2-2x-3x+3)-2\\=2x^2-5x+2-(2x^2-5x+3)-2\\=2x^2-5x+2-2x^2+5x-3-2\\=(2x^2-2x^2)+(-5x+5x)+(2-3-2)\\=-3\)
b)
\(x(x+3y+1)-2y(x-1)-(y+x+1)x\\=x^2+3xy+x-2xy+2y-xy-x^2-x\\=(x^2-x^2)+(3xy-2xy-xy)+(x-x)+2y\\=2y\)
Bài 2.
a)
\((14x^3+12x^2-14x):2x=(x+2)(3x-4)\\\Leftrightarrow 14x^3:2x+12x^2:2x-14x:2x=3x^2-4x+6x-8\\ \Leftrightarrow 7x^2+6x-7=3x^2+2x-8\\\Leftrightarrow (7x^2-3x^2)+(6x-2x)+(-7+8)=0\\\Leftrightarrow 4x^2+4x+1=0\\\Leftrightarrow (2x)^2+2\cdot 2x\cdot 1+1^2=0\\\Leftrightarrow (2x+1)^2=0\\\Leftrightarrow 2x+1=0\\\Leftrightarrow 2x=-1\\\Leftrightarrow x=\frac{-1}2\)
b)
\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)
15 tháng 11 2021
Bài 1:
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3a^2b-3ab^2-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\left(do.a+b+c\ne0\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)
\(M=\dfrac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\dfrac{3a^2}{\left(3a\right)^2}=\dfrac{3a^2}{9a^2}=\dfrac{1}{3}\)
15 tháng 11 2021
Bài 2:
a) \(=\dfrac{x\left(x^2+x-6\right)}{x\left(x^2-4\right)}=\dfrac{x\left(x-2\right)\left(x+3\right)}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x+3}{x+2}\)
b) \(=\dfrac{x\left(x+1\right)+7\left(x+1\right)}{x\left(x^2+2x+1\right)}=\dfrac{\left(x+1\right)\left(x+7\right)}{x\left(x+1\right)^2}=\dfrac{x+7}{x\left(x+1\right)}=\dfrac{x+7}{x^2+x}\)