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a) \(4x^2-12x=-9\)
\(\Leftrightarrow4x^2-12x+9=0\)
\(\Leftrightarrow\left(2x-3\right)^2=0\)
\(\Leftrightarrow2x-3=0\Leftrightarrow x=\frac{3}{2}\)
b) \(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(25-4x^2\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)+\left(5-2x\right)\left(5+2x\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7+5+2x\right)=0\)
\(\Leftrightarrow\left(5-2x\right)\left(4x+12\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-3\end{array}\right.\)
c)\(x^3+27+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow\left(x+3\right)x\left(x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-3\\x=0\\x=2\end{array}\right.\)
d) \(4\left(2x+7\right)^2-9\left(x+3\right)^2=0\)
\(\Leftrightarrow\left[2\left(2x+7\right)-3\left(x+3\right)\right]\left[2\left(2x+7\right)+3\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(4x+14-3x-9\right)\left(4x+14+3x+9\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-5\\x=-\frac{23}{17}\end{array}\right.\)
câu a, b, c dễ mà. Bạn áp dụng 7 hằng đẳng thúc là làm đc thoii!!
vd: a) \(\left(9x^2-4\right)\left(x+1\right)=\left(3x+2\right)\left(x^2-1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=\left(3x+2\right)\left(x-1\right)\left(x+1\right)\)
\(\Rightarrow\left(3x-2\right)\left(3x+2\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)[\left(3x-2\right)-\left(x-1\right)]=0\)
\(\Rightarrow\left(3x+2\right)\left(x+1\right)\left(2x-1\right)=0\) (bạn phá ngoặc ra rồi tính là ra bước này)
\(\Leftrightarrow3x+2=0\) hoặc \(x+1=0\) hoặc \(2x-1=0\) ( đến đây bạn chia làm 3 trường hợp r tự tính nhé)
Chúc bạn học tốt!!
d/
\(\Leftrightarrow x^3\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x^3+1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
e/
\(\Leftrightarrow x^3+x^2-6x-x^2-x+6=0\)
\(\Leftrightarrow x\left(x^2+x-6\right)-\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=-3\end{matrix}\right.\)
\(\left(x-1\right)^2-1+x^2=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow\left(x-1\right)^2+\left(x-1\right)\left(x+1\right)=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow2x\left(x-1\right)=\left(1-x\right)\left(x+3\right)\)
\(\Leftrightarrow2x\left(x-1\right)+\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+3\right)=0\)
\(\Rightarrow x=\pm1\)
Giúp tớ mấy câu còn lại đi các cậu, tớ cần gấp lắm ạ ;;-;;
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)\(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\left(x+2\right)\)
\(\Leftrightarrow x^3-3x^2+3x-1+3\left(x^2+2x+1\right)=x^3+8\)
\(\Leftrightarrow-3x^2+3x+3x^2+6x+3=9\)
\(\Leftrightarrow9x=6\Leftrightarrow x=\frac{2}{3}\)
b) \(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow x^2-4=8x-16\)
\(\Leftrightarrow x^2-8x+12=0\)
\(\Leftrightarrow x^2-2x-6x+12=0\)
\(\Leftrightarrow x\left(x-2\right)-6\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=2\end{cases}}\)
c) \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)^2=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)^2-9\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-11=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=11\end{cases}}\)
d) \(4x^2-12x+9=\left(5-x\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2=\left(5-x\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}2x-3=5-x\\2x-3=x-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=-2\end{cases}}\)