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x^2+7x+10=0
x(x+7)=-10
=>x>0 x<0
x+7<0 x+7<0
Mà x+7>x
=>x<0 =>x<0
x+7>0 x>-7
=>x thuộc -1;-2;-3;-4;-5;-6
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x+4=0\Leftrightarrow x=-\dfrac{4}{3}\\ 2x\left(x-1\right)-\left(1+2x\right)=-34\\ \Leftrightarrow2x^2-2x-1-2x=-34\\ \Leftrightarrow2x^2-4x+33=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+30=0\\ \Leftrightarrow2\left(x-1\right)^2+30=0\\ \Leftrightarrow x\in\varnothing\left[2\left(x-1\right)^2+30\ge30>0\right]\\ x^2+9x-10=0\\ \Leftrightarrow x^2-x+10x-10=0\\ \Leftrightarrow\left(x-1\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-10\end{matrix}\right.\\ \left(7x-1\right)\left(2+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7x-1=0\\2+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)Ta có:
\(x^2-7x+12=0\)
\(\Leftrightarrow x^2-3x-4x+12=0\)
\(\Leftrightarrow x\left(x-3\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=3\end{matrix}\right.\)
b) Ta có:
\(x\left(x-4\right)-3\left(4-x\right)=0\)
\(\Leftrightarrow x\left(x-4\right)+3\left(x-4\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)=\left(x^2-7x+10\right)\left(x^2-7x-10\right)\)
Đặt \(x^2-7x=t\),khi đó:
\(C=\left(t+10\right).\left(t-10\right)=t^2-10^2=t^2-100\)
Vì \(t^2\ge0=>t^2-100\ge-100\) (với mọi t)
Dấu "=" xảy ra\(< =>t=0< =>x^2-7x=0< =>x\left(x-7\right)=0< =>\orbr{\begin{cases}x=0\\x=7\end{cases}}\)
Vậy minC=-100 khi x=0 hoặc x=7
![](https://rs.olm.vn/images/avt/0.png?1311)
c: ta có: \(7x^2-2x-5=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{5}{7}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(y^2\)+ 2xy-7x-12=0 <=> 4\(y^2\)+8xy-28x-48=0 <=> 4\(y^2\)-49+4x(2y-7)=-1
<=> (2y-7)(2y+7+4x)=-1
=> Ta có : 2y-7= -1 và 2y+7+4x= 1
hoặc 2y-7=1 và 2y+7+4x=-1
*) 2y-7=1và 2y+7+4x=-1 *) 2y-7=-1 và 2y+7+4x=1
=> x=-4 và y=4 =>x=-3 và y=3
Vậy x=-4 và y=4 Hoặc x=-3 và y=3
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-5x-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
Vậy....
\(2x\left(x+6\right)=7x+42\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)
Vậy......
\(x^3-5x^2+x-5=0\)
\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\)
\(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy...
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Đặt x^2+2x+2=t
\(\frac{4}{t-1}+\frac{3}{t+1}=\frac{3}{2}\Leftrightarrow\frac{4t+4+3t-3}{t^2-1}=\frac{7t+1}{t^2-1}=\frac{3}{2}\)
\(\Leftrightarrow14t+2=3t^2-3\Leftrightarrow3t^2-14t-5=3t\left(t-5\right)+t-5=0\)\(\Leftrightarrow\left(t-5\right)\left(3t+1\right)=0\Rightarrow\left[\begin{matrix}t=5\\t=-\frac{1}{3}\left(loai\right)\end{matrix}\right.\)
Với t=5 ta có (x+1)^2=4\(\Rightarrow\left[\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
Ta có:
x2 + 7x + 10 = 0
<=> x^2 + 5x + 2x + 10 = 0
<=> x(x + 5) + 2(x + 5) = 0
<=> (x+2)(x+5) = 0
<=> x+2=0 hoặc x+5=0
<=> x= -2 hoặc x= -5
Vậy x = -2; -5.
\(x^2+7x+10=0\)
\(\Leftrightarrow\left(x^2+5x\right)+\left(2x+10\right)=0\)
\(\Leftrightarrow x\left(x+5\right)+2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x+5=0\\x+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\x=-2\end{cases}}}\)