Giúp mình với ạ , nghĩ hoài mà không biết làm 😭😭
Bài 1
4 x ( 2 - | X | ) + 5 | X | = 7
Bài 2
| 3x -2 |^2004 = | 3x -2 |^2004
^ là mũ ạ
Bài 3
| 1 - 2x | + x + 2 = 0
Bài 4
| 5x -3 | = | 7 - x |
Bài 5
2^5x : 2^3x = 4
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Bài 1 :
a) \(3x\left(5x^2-2x-1\right)=3x\cdot5x^2+3x\left(-2x\right)+3x\left(-1\right)\)
\(=15x^3-6x^2-3x\)
b) \(\left(x^2-2xy+3\right)\left(-xy\right)\)
\(=x^2\left(-xy\right)-2xy\left(-xy\right)+3\left(-xy\right)\)
\(=-x^3y+2x^2y^2-3xy\)
c) \(\frac{1}{2}x^2y\left(2x^3-\frac{2}{5}xy-1\right)\)
\(=\frac{1}{2}x^2y\cdot2x^3+\frac{1}{2}x^2y\cdot\left(-\frac{2}{5}xy\right)+\frac{1}{2}x^2y\left(-1\right)\)
\(=x^5y-\frac{1}{5}x^3y^2-\frac{1}{2}x^2y\)
d) \(\frac{1}{2}xy\left(\frac{2}{3}x^2-\frac{3}{4}xy+\frac{4}{5}y^2\right)\)
\(=\frac{1}{2}xy\cdot\frac{2}{3}x^2+\frac{1}{2}xy\cdot\left(-\frac{3}{4}xy\right)+\frac{1}{2}xy\cdot\frac{4}{5}y^2\)
\(=\frac{1}{3}x^3y-\frac{3}{8}x^2y^2+\frac{2}{5}xy^3\)
e) \(\left(x^2y-xy+xy^2+y^3\right)\left(3xy^3\right)\)
= \(x^2y\cdot3xy^3-xy\cdot3xy^3+xy^2\cdot3xy^3+y^3\cdot3xy^3\)
\(=3x^3y^4-3x^2y^4+3x^2y^5+3xy^6\)
Bài 2 :
3(2x - 1) + 3(5 - x) = 6x - 3 + 15 - x = (6x - x) - 3 + 15 = 5x - 3 + 15
Thay x = -3/2 vào biểu thức trên ta có : \(5\cdot\left(-\frac{3}{2}\right)-3+15\)
\(=-\frac{15}{2}-3+15=\frac{9}{2}\)
b) 25x - 4(3x - 1) + 7(5 - 2x)
= 25x - 12x + 4 + 35 - 14x
= (25x - 12x - 14x) + 4 + 35 = -x + 4 + 35 = -x + 39
Thay \(x=2\)vào biểu thức trên ta có : -2 + 39 = 37
c) 4x - 2(10x + 1) + 8(x - 2)
= 4x - 20x - 2 + 8x - 16
= (4x - 20x + 8x) - 2 - 16 = -8x - 2 - 16 = -8x - 18
Thay x = 1/2 vào biểu thức trên ta có \(-8\cdot\frac{1}{2}-18=-4-18=-22\)
d) Tương tự
Bài 3:
a) \(2x\left(x-4\right)-x\left(2x+3\right)=4\)
=> 2x2 - 8x - 2x2 - 3x = 4
=> (2x2 - 2x2) + (-8x - 3x) = 4
=> -11x = 4
=> x = \(-\frac{4}{11}\)
b) x(5 - 2x) + 2x(x - 7) = 18
=> 5x - 2x2 + 2x2 - 14x = 18
=> 5x - 14x = 18
=> -9x = 18
=> x = -2
Còn 2 câu làm tương tự
Bài 7 :
\(\frac{1}{4}-\left(2x-1\right)^2=0\)
\(\left(2x-1\right)^2=\frac{1}{4}-0\)
\(\left(2x-1\right)^2=\frac{1}{4}\)
\(\left(2x-1\right)^2=\left(\frac{1}{2}\right)^2\)
TH1:\(\Rightarrow2x-1=\frac{1}{2}\)
\(2x=\frac{1}{2}+1\)
\(2x=\frac{3}{2}\)
\(x=\frac{3}{4}\)
TH2:\(\Rightarrow2x-1=-\frac{1}{2}\)
\(2x=-\frac{1}{2}+1\)
\(2x=\frac{1}{2}\)
\(x=\frac{1}{4}\)
Vậy x \(\in\left\{\frac{1}{4};\frac{3}{4}\right\}\)
Bài 6 :
\(3^{x+1}=81\)
\(3^{x+1}=3^4\)
\(x+1=4\)
\(\Rightarrow x=3\)
Vậy x = 3
a)\(x\left(x-3\right)-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
b)\(\left(3x-5\right)\left(5x-7\right)+\left(5x+1\right)\left(2-3x\right)=4\)
\(\Leftrightarrow15x^2-46x+35-15x^2+7x+2-4=0\)
\(\Leftrightarrow33-39x=0\Leftrightarrow33=39x\Leftrightarrow x=\frac{33}{39}\)
a) \(x\left(x-3\right)-2x+6=0\)
\(x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}}\)
b) \((3x-5)(5x-7)+(5x+1)(2-3x)=4\)
\(15x^2-46x+35+10x-15x^2+2-3x-4=0\)
\(33-39x=0\)
\(3\left(11-13x\right)=0\)
\(11-13x=0\)
\(13x=11\)
\(x=\frac{11}{13}\)
a)
5.(12-x)-20=30
⇒60-5x-20=30
⇒-5x=30+20-60
⇒-5x=-10
⇒x=2
b)(17x - 25 ) : 8 + 65 = 92
(17x - 25 ) : 8 + 65 = 81
17x - 25 = 16 x 8 = 128
17x = 128+25=153
x= 153:17 =9
c)
x=23
Giải thích các bước giải:
3x – 10 = 2x + 13
3x-2x=13+10
x=23
d)4(2x+7)-3(3x-2)=24
4.2x+4.7-3.3x+3.2=24
8x+28-9x+6=24
8x-9x=24-28-6=-10
=>(-1)x=-10
x=-10:(-1)
x=10
a. \(5\cdot\left(12-x\right)-20=30\Leftrightarrow5\left(12-x\right)=50\)
\(\Leftrightarrow12-x=50:5=10\)
\(\Leftrightarrow x=12-10=2\)
b. \(\left(17x-25\right):8+65=9^2\)
\(\Leftrightarrow\left(17x-25\right):8=81-65=16\)
\(\Leftrightarrow17x-25=16:8=2\)
\(\Leftrightarrow17x=2+25=27\Leftrightarrow x=\frac{27}{17}\)
c. \(3x-10=2x+13\)
\(\Leftrightarrow3x-2x=10+13\)
\(\Leftrightarrow x=23\)
d. \(4\cdot\left(2x+7\right)-3\cdot\left(3x-2\right)=24\)
\(\Leftrightarrow8x+28-9x+6=24\)
\(\Leftrightarrow34-x=24\Leftrightarrow x=10\)
`@` `\text {Ans}`
`\downarrow`
`(8x-3)(3x+2)-(4x+7)(x+4)=(2x+1)(5x-1)-33`
`\Leftrightarrow 8x(3x+2) -3(3x+2) - 4x(x+4) + 7(x+4) = 2x(5x-1) + 5x-1 - 33`
`\Leftrightarrow 24x^2 + 16x - 9x - 6 - 4x^2 - 16x - 7x - 28 = 10x^2 - 2x + 5x - 1 - 33`
`\Leftrightarrow 20x^2 -16x - 34 = 10x^2 + 3x - 34`
`\Leftrightarrow 20x^2 - 16x - 34 - 10x^2 - 3x + 34 = 0`
`\Leftrightarrow 10x^2 - 19x = 0`
`\Leftrightarrow x(10x - 19)=0`
`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\10x-19=0\end{matrix}\right.\)
`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\10x=19\end{matrix}\right.\)
`\Leftrightarrow `\(\left[{}\begin{matrix}x=0\\x=\dfrac{19}{10}\end{matrix}\right.\)
Vậy, `x={0; 19/10}.`
1, \(3x\left(x-7\right)+2x-14=0\)
\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)
\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)
2, \(x^3+3x^2-\left(x+3\right)=0\)
\(\Rightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)
3, \(15x-5+6x^2-2x=0\)
\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)
\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)
4, \(5x-2-25x^2+10x=0\)
\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)
\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)
\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)
Bài 3:
\(\left|1-2x\right|+x+2=0\)
⇒ \(\left|1-2x\right|+x=0-2\)
⇒ \(\left|1-2x\right|+x=-2\)
⇒ \(\left|1-2x\right|=-2-x\)
⇒ \(\left[{}\begin{matrix}1-2x=-2-x\\1-2x=2+x\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}1+2=-x+2x\\1-2=x+2x\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}3=1x\\-1=3x\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=3:1\\x=\left(-1\right):3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=3\\x=-\frac{1}{3}\end{matrix}\right.\)
Vậy \(x\in\left\{3;-\frac{1}{3}\right\}.\)
Bài 4:
\(\left|5x-3\right|=\left|7-x\right|\)
⇒ \(\left[{}\begin{matrix}5x-3=7-x\\5x-3=x-7\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}5x+x=7+3\\5x-x=\left(-7\right)+3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}6x=10\\4x=-4\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=10:6\\x=\left(-4\right):4\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\frac{5}{3}\\x=-1\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{5}{3};-1\right\}.\)
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