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1, \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)
\(\Leftrightarrow-4x^2+28x+4x^3-20x=28x^2-13\)
\(\Leftrightarrow-32x^2+8x+4x^3-13=0\)( vô nghiệm )
2, \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)
\(\Leftrightarrow12x^3-7x^2-10x-7x^2-35x=-2x^2+11x-12+12x^3+2x^2\)
\(\Leftrightarrow12x^3-14x^2-45x=11x-12+12x^3\)
\(\Leftrightarrow-14x^2-56x-12=0\)( vô nghiệm )
Mình làm riêng ra nhá , chứ nhiều quá nên thông cảm cho mình :))
1. \(-4x\left(x-7\right)+4x\left(x^2-5\right)=28x^2-13\)
=> \(-4x^2+28x+4x^3-20x=28x^2-13\)
=> \(-4x^2+4x^3+\left(28x-20x\right)=28x^2-13\)
=> \(-4x^2+4x^3+8x-28x^2+13=0\)
=> \(\left(-4x^2-28x^2\right)+4x^3+8x+13=0\)
=> \(-32x^2+4x^3+8x+13=0\)
=> vô nghiệm
2. \(\left(4x^2-5x\right)\left(3x+2\right)-7x\left(x+5\right)=\left(-4+x\right)\left(-2x+3\right)+12x^3+2x^2\)
=> \(4x^2\left(3x+2\right)-5x\left(3x+2\right)-7x\left(x+5\right)=-4\left(-2x+3\right)+x\left(-2x+3\right)+12x^3+2x^2\)
=> \(12x^3+8x^2-15x^2-10x-7x^2-35x=8x-12-2x^2+3x+12x^3+2x^2\)
=> \(12x^3+8x^2-15x^2-10x-7x^2-35x-8x+12+2x^2-3x-12x^3-2x^2=0\)
=> \(\left(12x^3-12x^3\right)+\left(8x^2-15x^2-7x^2+2x^2-2x^2\right)+\left(-10x-35x-8x-3x\right)+12=0\)
=> \(-14x^2-56x+12=0\)
=> .... tự tìm
Câu c dấu bằng chỗ nào ?
a) \(\left(2x-1\right)\left(2x+1\right)-4x^2=3\Leftrightarrow\left(4x^2-1\right)-4x^2=3\Rightarrow-1=3\) (không đúng)
Tí làm tiếp nhé ;) h đi chơi đã
a) 5x ( x - 3 )2 - 5 ( x - 1 )3 + 15 ( x + 2 )( x - 2) = 5
⇔ x ( x - 3 )2 - ( x - 1 )3 + 3 ( x + 2 ) ( x - 2 ) = 1
⇔ x (x2 - 6x + 9) - (x3 - 3x2 + 3x - 1) + 3(x2 - 4) = 1
⇔ x3 - 6x2 + 9x - x3 + 3x2 - 3x + 1 + 3x2 - 12 = 1
⇔ 6x - 11 = 1 ⇔ 6x = 12 ⇔ x = 2
b) ( x + 2 ) ( 3 - 4x ) = x2 + 4x + 4
⇔ ( x + 2 ) ( 3 - 4x ) = ( x + 2 )2
⇔ 3 - 4x = x + 2 ⇔ 5x = 1
⇔ x = 1/5
b) \(\left(x+2\right).\left(3-4x\right)=x^2+4x+4\)
\(\Rightarrow\left(x+2\right).\left(3-4x\right)=x^2+2.x.2+2^2\)
\(\Rightarrow\left(x+2\right).\left(3-4x\right)=\left(x+2\right)^2\)
\(\Rightarrow\left(3-4x\right)=\left(x+2\right)^2:\left(x+2\right)\)
\(\Rightarrow3-4x=x+2\)
\(\Rightarrow3-2=x+4x\)
\(\Rightarrow1=5x\)
\(\Rightarrow x=1:5\)
\(\Rightarrow x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}.\)
Chúc bạn học tốt!
\(\left(5x-1\right)^2-\left(5x-4\right)\left(5x+4\right)=7\)
\(25x^2-10x+1-25x^2+16=7\)
\(17-10x=7\)
\(10x=10\)
\(x=1\)
a) \(\left(2x+3\right)\left(x-4\right)+\left(x+5\right)\left(x-2\right)=\left(3x-5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x+10=3x^2-12x+20\)
\(\Leftrightarrow3x^2-7x-2=3x^2-12x+20\)
\(\Leftrightarrow-7x+12x=20+2\)
\(\Leftrightarrow5x=22\)
\(\Rightarrow x=\dfrac{22}{5}\)
tick cho mk nha
b) \(\left(8x-3\right)\left(3x+2\right)-\left(4x+7\right)\left(x+4\right)=\left(2x+1\right)\left(5x-1\right)\)
\(\Leftrightarrow24x^2+16x-9x-6-4x^2-23x-28=10x^2+3x-1\)
\(\Leftrightarrow20x^2-16x-34-10x^2-3x+1=0\)
\(\Leftrightarrow10x^2-19x-33=0\)
\(\Delta=\left(-19\right)^2-4.10.\left(-33\right)=1320\)
\(x_1=3;x_2=\dfrac{-11}{10}\)
Tick cho mk nha
a)
\(x\left(x-2\right)-x+2=0\\ \Leftrightarrow-x\left(2-x\right)+\left(2-x\right)=0\\ \Leftrightarrow\left(2-x\right)\left(1-x\right)=0\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2-x=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
Vậy...
b)
\(x^2\left(x^2+1\right)-x^2-1=0\\ \Leftrightarrow x^2\left(x^2+1\right)-\left(x^2+1\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\\x^2-1=0\end{matrix}\right.\)
Vì \(x^2+1\) luôn lớn hơn 0 với mọi x
\(\Rightarrow x^2-1=0\\ \Leftrightarrow x=1\)
Vậy...
c)
\(5x(x-3)^2-5(x-1)^3+15(x+2)(x-2)=5\)
\(\Leftrightarrow x(x-3)^2-(x-1)^3+3(x+2)(x-2)=1\)
\(\Leftrightarrow x(x^2-6x+9)-(x^3-3x^2+3x-1)+3(x^2-4)=1\)
\(\Leftrightarrow 6x-12=0\Rightarrow x=2\)
_-_