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a)\(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)
\(4x^2-20x-\left(4x^2-7x+3\right)=5\)
\(4x^2-20x-4x^2+7x-3=5\)
\(-13x=8\)
\(x=-\frac{8}{13}\)
b)\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(48x^2-32x+5+3x-48x^2-7+112x=81\)
\(83x-2=81\)
\(x=1\)
\(\left(12x-5\right)\left(4x-1\right)\left(3x-7\right)\left(1-16x\right)=81\)
\(\Rightarrow\left(48x^2-12x-20x+5\right)\left(3x-7\right)\left(1-16x\right)=81\)
\(\Rightarrow\left(48x^2-32x+5\right)\left(3x-7\right)\left(1-16x\right)=81\)
\(\Rightarrow\left(144x^3-336x^2-96x^2+224x+15x-35\right)\left(1-16x\right)=81\)
\(\Rightarrow\left(144x^3-432x^2+239x-35\right)\left(1-16x\right)=81\)
\(\Rightarrow144x^3-2304x^4-432x^2+6912x^3+239x-3824x^2-35+560x=81\)
\(\Rightarrow-2304x^4+7056x^3-4256x^2+799x-116=0\)
\(\Rightarrow\left[{}\begin{matrix}x_1\approx2,3\\x_2\approx0,5\end{matrix}\right.\)
Bài 1:
a) \(-5\left(x^2-3x+1\right)+x\left(1+5x\right)=x-2\)
\(\Rightarrow-5x^2+15x-5+x+5x^2=x-2\)
\(\Rightarrow16x-5=x-2\)
\(\Rightarrow16x-x=5-2\)
\(\Rightarrow15x=3\)
\(\Rightarrow x=\dfrac{15}{3}=5\)
b) \(12x^2-4x\left(3x+5\right)=10x-17\)
\(\Rightarrow12x^2-12x^2-20x=10x-17\)
\(\Rightarrow-20x=10x-17\)
\(\Rightarrow-20x-10x=-17\)
\(\Rightarrow-30x=-17\)
\(\Rightarrow x=\dfrac{-30}{-17}=\dfrac{30}{17}\)
c) \(-4x\left(x-5\right)+7x\left(x-4\right)-3x^2=12\)
\(\Rightarrow-4x^2+20x+7x^2-28x-3x^2=12\)
\(\Rightarrow-8x=12\)
\(\Rightarrow x=\dfrac{12}{-8}=-\dfrac{4}{3}\)
Bài 2:
a) \(\left(x+5\right)\left(x-7\right)-7x\left(x-3\right)\)
\(=x^2-7x+5x-35-7x^2+21x\)
\(=-6x^2+19x-35\)
b) \(x\left(x^2-x-2\right)-\left(x-5\right)\left(x+1\right)\)
\(=x^3-x^2-2x-x^2+x-5x-5\)
\(=x^3-2x^2-6x-5\)
c) \(\left(x-5\right)\left(x-7\right)-\left(x+4\right)\left(x-3\right)\)
\(=x^2-7x-5x+35-x^2-3x+4x-12\)
\(=11x+23\)
d) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)
\(=x^2-2x-x+2-x^2+2x+5x+10\)
\(=4x+12\)
\(\left(4x-1\right)^3+\left(3-4x\right)\left(9+12x+16x\right)=\left(8x-1\right)\left(8x+1\right)-\left(3x-5\right)\)
\(< =>64x^3-3x^2+3x-1+\left(3x^2-4^3\right)=64x^2-1-3x+5\)
\(< =>64x^3+\left(3x^2-3x^2\right)+3x-\left(1+64\right)=64x^2-3x+4\)
\(< =>64x^3+3x-65-64x^2+3x-4=0\)
\(< =>64x^3-64x^2+6x-69=0\)
số to nên mình lười cardano , nên bạn xét vô nghiệm cũng được
phát hiện lỗi sai của mình rồi , mình xin lỗi nhé
từ dòng 2 trở đi : \(< =>64x^3-48x^2+12x-1+\left(3^3-64x^3\right)=64x^2-3x+4\)
\(< =>64x^3-64x^3-48x^2-64x^2+12x+26+3x-4\)
\(< =>-112x^2+15x+22=0\)
Bạn dùng máy tính hoặc đen ta cũng được nhé
#)Góp ý :
Bạn ghi thiếu đề phải không ? (12x-5)(4x-1)+(3x-7)(1-16x) = 81
#)Giải :
\(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
\(\Leftrightarrow12x\left(4x-1\right)-5\left(4x-1\right)+3x\left(1-16x\right)-7\left(1-16x\right)=81\)
\(\Leftrightarrow12x.4x-12x-20x+5-3x+3x.16x-7+7.16x=81\)
\(\Leftrightarrow48x^2-12x-12x-20+5+3x-48x^2-7+112x=81\)
\(\Leftrightarrow\left(48x^2-48x^2\right)+\left(112x-12x-20x+3x\right)-\left(7-5\right)=81\)
\(\Leftrightarrow83x=81+2\)
\(\Leftrightarrow x=1\)
a/ \(\left(x-1\right)\left(x^2+x+1\right)=-9\)
=> \(x^2\left(x-1\right)+x\left(x-1\right)+\left(x-1\right)=-9\)
=> \(x^3-x^2+x^2-x+x-1=-9\)
=> \(x^3+\left(x^2-x^2\right)+\left(x-x\right)-1=-9\)
=> \(x^3-1=-9\)
=> \(x^3=-8\)
=> \(x=-2\)
b/ \(\left(12x-5\right)\left(4x-1\right)+\left(3x-7\right)\left(1-16x\right)=81\)
=> \(48x^2-12x-20x+5+3x-48x^2-7+112x=81\)
=> \(\left(48x^2-48x^2\right)+\left(112x-12x-20x+3x\right)+\left(5-7\right)=81\)
=> \(83x-2=81\)
=> \(83x=83\)
=> x = 1