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1) \(\left(x+1\right)\left(x+2\right)-3x\left(x-4\right)=x^2+3x+2-3x^2+12x=-2x^2+15x+2\)
2) \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)\)
\(\Leftrightarrow3x^2-10x+8=3x^2-27x\)
\(\Leftrightarrow17x=-8\Leftrightarrow x=-\dfrac{8}{17}\)
3) \(-3\left(x-4\right)\left(x-2\right)-x^2\left(-3x+18\right)+24x-25\)
\(=-3x^3+6x^2+12x^2-24x+3x^3-18x^2+24x-25=-25\)
\(\left(3x-4\right)^2-2\left(3x-4\right)\left(x-4\right)+\left(x-4\right)^2\)
\(=\left(3x-4-x+4\right)^2\)
\(=4x^2\)
\(\left(3x-4\right)^2+\left(4-x\right)^2-2\left(3x-4\right)\left(x-4\right)=\left(3x-4\right)^2-2\left(3x-4\right)\left(x-4\right)+\left(x-4\right)^2=\left(3x-4-x+4\right)^2=\left(2x\right)^2=4x^2\)
a) (x2-1)(x2+4)(x2-4)=(x2-1)(x4-16)
b) 9x2+6x+1+4-9x2= 6x+5
\(\left(x+2\right)^2-\left(x+4\right)^2+x^2-3x+1\)
\(=x^2+4x+4-x^2-8x-16+x^2-3x+1\)
\(=x^2-7x-11\)
\(\left(x+2\right)^2-\left(x+4\right)^2+x^2-3x+1\)
\(=x^2+4x+4-x^2-8x-16+x^2-3x+1=x^2-7x-11\)
a: Ta có: \(x^2-4x\left(3x-4\right)+7x-5\)
\(=x^2-12x^2+16x+7x-5\)
\(=-11x^2+23x-5\)
b: Ta có: \(7x\left(x^2-5\right)-3x^2y\left(xy-6y^2\right)\)
\(=7x^3-35x-3x^3y^2+18x^2y^3\)
c: Ta có: \(\left(5x+4\right)\left(2x-7\right)\)
\(=10x^2-35x+8x-28\)
\(=10x^2-27x-28\)
\(A=x\left(9x^2-16\right)-9\left(x^3+8\right)+16x\\ A=9x^3-16x-9x^3-72+16x\\ A=-72\)
\(A=x\left(3x-4\right)\left(3x+4\right)-9\left(x+2\right)\left(x^2-2x+4\right)+16x\)
\(=x\left(9x^2-16\right)-9\left(x^3+8\right)+16x\)
\(=9x^3-16x-9x^3-72+16x=-72\)
a: =12x^3y^2-12x^3y^3+6x^2y^2
b: =\(\left(-3x+2\right)\left(5x^2-\dfrac{1}{3}x+4\right)\)
=-15x^3+x^2-12x+10x^2-2/3x+8
=-15x^3+11x^2-38/3x+8
c: =x^2-x-2+3x-x^2
=2x-2
a: Ta có: \(A=\dfrac{3x^2-12x+12}{x^2-4}\)
\(=\dfrac{3\left(x^2-4x+4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3x-6}{x+2}\)
b: Thay \(x=-\dfrac{1}{2}\) vào A, ta được:
\(A=\left(3\cdot\dfrac{-1}{2}-6\right):\left(-\dfrac{1}{2}+2\right)\)
\(=\left(-\dfrac{3}{2}-6\right):\dfrac{3}{2}\)
\(=\dfrac{-15}{2}\cdot\dfrac{2}{3}=-5\)
a: \(=\dfrac{\left(x-1\right)\left(x+1\right)}{\left(x+1\right)^2}=\dfrac{x-1}{x+1}\)
b: \(=\dfrac{\left(x-1\right)\cdot\left(x+1\right)}{\left(x-1\right)^2}=\dfrac{x+1}{x-1}\)
c: \(=\dfrac{3x+2-3x+2+3x-6}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{3x-2}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{1}{3x+2}\)
Bạn lưu ý viết đề bằng công thức toán để được hỗ trợ tốt hơn. Viết thế này nhìn khá khó đọc.
Để viết công thức toán bạn nhấn biểu tượng $\sum$ góc trái khung soạn thảo.
\(=\left(3x-4+4-x\right)^2=\left(2x\right)^2=4x^2\)
\(\left(3x-4\right)^2+2\left(3x-4\right)\left(x-4\right)+\left(x-4\right)^2\)
\(=\left(3x-4+x-4\right)^2\)
\(=\left(4x-8\right)^2\)