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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 ?
1. (3x - 5)2 - (3x + 1)2 = 8
=> (3x - 5 - 3x - 1)(3x - 5 + 3x + 1) = 8
=> -6(6x - 4) = 8
=> 6x - 4 = \(\dfrac{-4}{3}\)
\(\Rightarrow x=\dfrac{4}{9}\)
2) 2x(8x - 3) - (4x - 3)2 = 27
=> 16x2 - 6x - 16x2 + 24x - 9 = 27
=> 18x - 9 = 27
=> x = 2
3) (2x - 3)2 - (2x + 1)2 = 3
=> (2x - 3 - 2x - 1)(2x - 3 + 2x +1) = 3
=> -4(4x - 2) = 3
=> 4x - 2 = \(\dfrac{-3}{4}\)
\(\Rightarrow x=\dfrac{5}{16}\)
4) (x + 5)2 - x2 = 45
=> (x + 5 - x)(x + 5 + x) = 45
=> 5(2x + 5) = 45
=> 2x + 5 = 9
=> x = 2
5) (x - 3)3 - (x - 3)(x2 + 3x + 9) + 9(x + 1)2 = 18
=> x3 - 9x2 + 27x - 27 - x3 + 27 + 9(x2 + 2x + 1) = 18
=> -9x2 + 27x + 9x2 + 18x + 9 = 18
=> 45x + 9 = 18
=> 45x = 9
=> x = \(\dfrac{1}{5}\)
6) x(x - 4)(x + 4) - (x - 5)(x2 + 5x + 25) = 13
=> x (x2 - 16) - (x3 - 125) = 13
=> x3 - 16x - x3 + 125 = 13
=> -16x = -112
=> x = 7.
Bài 3:
a) \(\left(x-6\right).\left(2x-5\right).\left(3x+9\right)=0\)
\(\Leftrightarrow\left(x-6\right).\left(2x-5\right).3.\left(x+3\right)=0\)
Vì \(3\ne0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\2x-5=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\2x=5\\x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=\frac{5}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{6;\frac{5}{2};-3\right\}.\)
b) \(2x.\left(x-3\right)+5.\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right).\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\frac{5}{2}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{3;-\frac{5}{2}\right\}.\)
c) \(\left(x^2-4\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x^2-2^2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2\right)-\left(x-2\right).\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left(x-2\right).\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{2;\frac{1}{3}\right\}.\)
Chúc bạn học tốt!
a) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=46\)
\(\Leftrightarrow16x^2-9-16x^2+40x-25=46\)
\(\Leftrightarrow40x=46+9+25=80\)
\(\Leftrightarrow x=2\)
b) \(\left(x+1\right)^3+2x-\left(x-1\right)^3-3\left[\left(x+1\right)^2+\left(x-1\right)^2\right]+5=0\)
\(=x^3+3x^2+3x+1+2x-x^3+3x^2-3x+1-3\left(x^2+2x+1+x^2-2x+1\right)+5=0\)
\(=6x^2+2x+2-3\left(2x^2+2\right)+5=0\)
\(\Leftrightarrow6x^2+2x+2-6x^2-6+5=0\)
\(\Leftrightarrow2x=-2+6-5=-1\)
\(\Leftrightarrow x=\frac{1}{2}\)
a) \(4x^2-8x=0\)
\(\Rightarrow4x\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=0+2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy \(x_1=0;x_2=2\)
b) \(\left(x+5\right)-3x\left(x+5\right)=0\)
\(\Rightarrow-3x^2-14x+5=0\)
\(\Leftrightarrow\left(-3x+1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-3x+1=0\\x+5=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=-5\end{matrix}\right.\)
Vậy \(x_1=-5;x_2=\dfrac{1}{3}\)
\(a,4x^2-8x=0\Rightarrow4x\left(x-8\right)=0\Rightarrow\left[{}\begin{matrix}4x=0\\x-8=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)\(b,\left(x+5\right)-3x\left(x+5\right)=0\Leftrightarrow\left(x+5\right)\left(1-3x\right)=0\Rightarrow\left[{}\begin{matrix}x+5=0\\1-3x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\3x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{1}{3}\end{matrix}\right.\)