38-2x=9
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sửa đề: \(B=2x+4\sqrt{x}+9\)ĐK : x >= 0
\(=2\left(x+2\sqrt{x}+1-1\right)+9=2\left(\sqrt{x}+1\right)^2+7\ge9\)
Dấu ''='' xảy ra khi x = 0
Vậy GTNN của B bằng 9 tại x = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+38\)
\(\Leftrightarrow\left[\left(3x+2\right)-\left(3x-2\right)\right]\left[\left(3x+2\right)+\left(3x-2\right)\right]=5x+38\)
\(\Leftrightarrow\left(3x+2-3x+2\right)\left(3x+2+3x-2\right)=5x+38\)
\(\Leftrightarrow4\cdot6x=5x+38\)
\(\Leftrightarrow24x-5x=38\)
\(\Leftrightarrow19x=38\Leftrightarrow x=\dfrac{38}{19}=2\)
Vậy \(S=\left\{2\right\}\)
b) \(\left(x+1\right)\left(x^2-2x+1\right)-2x=2\left(x-1\right)\left(x+1\right)\)
\(\Leftrightarrow x^3-2x^2+x+x^2-2x+1-2x=2\left(x^2-1\right)\)
\(\Leftrightarrow x^3-2x^2+x+x^2-2x+1-2x=2x^2-2\)
\(\Leftrightarrow x^3-2x^2+x+x^2-2x+1-2x-2x^2+2=0\)
\(\Leftrightarrow x^3-3x^2-3x+3=0\)
PT vô nghiệm , không tìm được x
Vậy \(S=\varnothing\)
c) \(3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\)
\(\Leftrightarrow3\left(x^2-2x+4\right)+9\left(x-1\right)=3\left(x^2+x-3\right)\)
\(\Leftrightarrow3x^2-6x+12+9x-9=3x^2+3x-9\)
\(\Leftrightarrow3x^2-6x+12+9x-9-3x^2-3x+9=0\)
\(\Leftrightarrow0x+12=0\)
PT vô nghiệm
Vậy \(S=\varnothing\)
Câu cuối tương tự
![](https://rs.olm.vn/images/avt/0.png?1311)
a,( x - 29 ) - ( 17 - 38 ) = -9
( x - 29 ) + 21 = - 9
x - 29 = - 9 - 21
x - 29 = - 30
x = -30 + 29
x = -1
c, ( 27 - x ) + ( 15 + x ) = - 24
27 - x + 15 + x = -24
- x + x = -24-27-15
Vô lí vì 0 ko bằng -66
Vậy \(x\in\varnothing\)
d, |2x - 7|- 9 =20
|2x-7|=11
* 2x-7=11 * 2x-7=-11
2x=11+7 2x=-11+7
2x=18 2x=-4
x=18:2 x=-4:2
x=9 x=-2
Vậy x=9 hoặc x=-2
![](https://rs.olm.vn/images/avt/0.png?1311)
1) Ta có: \(\left(-86-x\right)-\left(3+2x\right)=-4-15\)
\(\Leftrightarrow-86-x-3-2x+4+15=0\)
\(\Leftrightarrow-3x-70=0\)
\(\Leftrightarrow-3x=70\)
hay \(x=-\dfrac{70}{3}\)
Vậy: \(x=-\dfrac{70}{3}\)
2) Ta có: \(18+\left(-x\right)-\left(40-28\right)=-32-\left(-18\right)\)
\(\Leftrightarrow18-x-40+28+32-18=0\)
\(\Leftrightarrow-x+20=0\)
\(\Leftrightarrow-x=-20\)
hay x=20
Vậy: x=20
3) Ta có: \(-27-\left(-31+x\right)-25=-5-17\)
\(\Leftrightarrow-27+31-x-25+5+17=0\)
\(\Leftrightarrow-x+1=0\)
\(\Leftrightarrow-x=-1\)
hay x=1
Vậy: x=1
4) Ta có: \(-9-14-x+42-38=-5+13\)
\(\Leftrightarrow-x-19=8\)
\(\Leftrightarrow-x=27\)
hay x=-27
Vậy: x=-27
![](https://rs.olm.vn/images/avt/0.png?1311)
e: =>-40+3+33+40-x=47
=>36-x=47
=>x=-11
f: =>x(x-3)(11-x)(11+x)=0
hay \(x\in\left\{0;3;11;-11\right\}\)
g: =>-62-38-x+2x=-100
=>x-100=-100
hay x=0
![](https://rs.olm.vn/images/avt/0.png?1311)
i: =>x-12-2x-31=6
=>-x-43=6
=>x+43=-6
hay x=-49
h: =>(x+1)=0
=>x=-1
f: =>x(x-3)(x+11)(x-11)=0
hay \(x\in\left\{0;3;-11;11\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(2x+3⋮2x-1\)
\(2x-1+4⋮2x-1\)
\(4⋮2x+1\)hay \(2x+1\inƯ\left(4\right)=\left\{1;2;4\right\}\)
2x + 1 | 1 | 2 | 4 |
2x | 0 | 1 | 3 |
x | 0 | 1/2 | 3/2 |
c, \(\left(x+5\right)\left(y-3\right)=15\Leftrightarrow x+5;y-3\inƯ\left(15\right)=\left\{1;3;5;15\right\}\)
x + 5 | 1 | 3 | 5 | 15 |
y - 3 | 15 | 5 | 3 | 1 |
x | -4 | -2 | 0 | 10 |
y | 18 | 8 | 6 | 4 |
38-2x=9
2x=38-9
2x=29
x=29:2
x=14,5