Tìm x biết 11x + 5 = 60
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ( 2 + 1 )x = 45
3x = 45
x = 45 : 3
x = 15
b) ( 2 + 7 )x = 918
9x = 918
x = 918 : 9
x = 102
c) ( 2 + 3 )x = 65
5x = 65
x = 65 : 5
x = 13
d) ( 11 + 22 )x = 66
33 x = 66
x = 66 : 33
x = 2
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x^2+3x+2\right)\left(x^2+11x+30\right)-60=0\)
\(\Rightarrow\left[\left(x+1\right)\left(x+2\right)\right].\left[\left(x+5\right)\left(x+6\right)\right]-60=0\)
\(\Rightarrow\left[\left(x+1\right)\left(x+6\right)\right].\left[\left(x+2\right)\left(x+5\right)\right]-60=0\)
\(\Rightarrow\left(x^2+7x+6\right)\left(x^2+7x+10\right)-60=0\left(1\right)\)
Đặt \(x^2+7x+6=a\Rightarrow x^2+7x+10=a+4\)
Thay vào (1), ta có:
\(a\left(a+4\right)-60=0\)
\(\Rightarrow a^2+4a-60=0\)
\(\Rightarrow a^2+10a-6a-60=0\)
\(\Rightarrow a\left(a+10\right)-6\left(a+10\right)=0\)
\(\Rightarrow\left(a-6\right)\left(a+10\right)=0\)
\(\Rightarrow\orbr{\begin{cases}a=6\\a=-10\end{cases}}\)
- Nếu \(x^2+7x+6=6\)
\(\Rightarrow x^2+7x=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}\)
- Nếu \(x^2+7x+6=-10\)
\(\Rightarrow x^2+7x+16=0\)
Mà \(x^2+7x+16=x^2+2.x.\frac{7}{2}+\frac{49}{4}+\frac{15}{4}=\left(x+\frac{7}{2}\right)^2+\frac{15}{4}>0\forall x\)
Vậy \(x=0,x=-7\)
Học tốt.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=-11x^5+4x-12x^2+11x^5+13x^2-7x+2\)
\(A=\left(-11x^5+11x^5\right)+\left(-12x^2+13x^2\right)+\left(4x-7x\right)+2\)
\(A=0+x^2+\left(-3x\right)+2\)
\(A=x^2-3x+2\)
Bậc của đa thức là: \(2\)
Hệ số cao nhất là: \(1\)
b) Ta có: \(M\left(x\right)=A\left(x\right)\cdot B\left(x\right)\)
\(\Rightarrow M\left(x\right)=\left(x^2-3x+2\right)\cdot\left(x-1\right)\)
\(\Rightarrow M\left(x\right)=x^3-x^2-3x^2+3x+2x-2\)
\(\Rightarrow M\left(x\right)=x^3-4x^2+5x-2\)
c) A(x) có nghiệm khi:
\(A\left(x\right)=0\)
\(\Rightarrow x^2-3x+2=0\)
\(\Rightarrow x^2-x-2x+2=0\)
\(\Rightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, (2\(x\) -5) + 17 = 6
2\(x\) - 5 = 6 - 17
2\(x\) - 5 = - 11
2\(x\) = - 11 + 5
2\(x\) = -6
\(x\) = -3
b, (-18).\(x\) = - 36
\(x\) = -36 : (-18)
\(x\) = 2
c, 15 - (-11\(x\) - 7) = -22
- 11\(x\) - 7 = 15 + 22
-11\(x\) - 7 = 37
11\(x\) = -37 - 7
11\(x\) = - 44
\(x\) = - 4
a) \(\left(2x-5\right)+17=6\Rightarrow2x-5=-11\Rightarrow2x=-6\Rightarrow x=-3\)
b) \(\left(-18\right)x=-36\Rightarrow x=\left(-36\right):\left(-18\right)=2\)
c) \(15-\left(-11x-7\right)=-22\)
\(\Rightarrow15+11x+7=-22\)
\(\Rightarrow11x=-22-7-15\)
\(\Rightarrow11x=-44\)
\(\Rightarrow x=-44:11=-4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,3.\left(x-4\right)-\left(8-x\right)=12\)
\(3x-12-8+x=12\)
\(4x-20=12\)
\(4x=12+20\)
\(4x=32\)
\(x=32:4\)
\(x=8\)
\(b,4.\left(x-5\right)-\left(x+7\right)=-19\)
\(4x-20-x-7=-19\)
\(3x-27=-19\)
\(3x=-19+27\)
\(3x=8\)
\(x=\frac{8}{3}\)
\(c,7.\left(x-3\right)-5.\left(3-x\right)=11x-5\)
\(7x-21-15+5x=11x-5\)
\(12x-36=11x-5\)
\(12x-11x=-5+36\)
\(x=31\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x = 5