4x-2 x 16-47
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a) x- 18 = - 25
x =( -25 ) + 18
x = -7
b) 42 - x = 53 - 60
16 - x = 65
x = 16 - 65
x = -49
(25-32)-(7+x)=47-25
-7 - ( 7 + x ) = 22
-7 - 7 - x = 22
- 14 - x = 22
x = 22 + ( - 14 )
x = 8
3x - 16 = 40 + x
3x - x = 40 + 16
2x = 56
x = 56 : 2
x = 28
-4 . | 2x - 3 | = ( - 16 ) : | - 4 |
-4 . | 2x - 3 | = ( - 16 ) : 4
-4 . | 2x - 3 | = -4
| 2x -3 | = -4 : ( - 4 )
| 2x - 3 | = 1
=> \(\orbr{\begin{cases}2x-3=1\\2x=3=-1\end{cases}\Rightarrow\orbr{\begin{cases}2x=1+3=4\\2x=-1+3=2\end{cases}\Rightarrow}\orbr{\begin{cases}x=4:2=2\\x=2:2=1\end{cases}}}\)
Vậy x = 2 hoặc 1
25 – 2x = 16 – 3x
-2x + 3x = 16 - 25
x = -9
Vậy x = -9
x - (47 - 22 ) = 5 + (10-4x)
x - 47 + 22 = 5 + 10 - 4x
x + 4x = 5 + 10 - 22 + 47
5x = 40
x = 40 : 5
x = 8
Vậy x = 8.
~ HOK TỐT ~
\(\text{25 – 2x = 16 – 3x}\)
\(2x+3x=16-25\)
\(5x=-9\)
\(\Rightarrow x=\frac{-9}{5}\)
\(\text{d) x – (47 – 22) = 5+ (10 – 4x)}\)
\(x-47+22=5+10-4x\)
\(x+4x=5+10-22+47\)
\(5x=40\)
\(\Rightarrow x=8\)
học tốt
a, ( 1+x )^3 = (2x)^3
b, ( x-1 )^2=16
c, (x+1)^2=25
d, 4x^3+15=47
e,(2x-1)^5=x^5
Mn giải nhanh giúp mk vs
a,\(\left(1+x\right)^3=\left(2x\right)^3\)
=>\(1+x=2x\)
=>\(x-2x=-1\)
=>\(-x=-1\)
=>\(x=1\)
vậy \(x=1\)
b,\(\left(x-1\right)^2=16\)
=>\(\left(x-1\right)^2=4^2\)
=>\(x-1=4\)
=>\(x=4+1\)
=>\(x=5\)
Vậy\(x=5\)
c,\(\left(x+1\right)^2=25\)
=>\(\left(x+1\right)^2=5^2\)
=>\(x+1=5\)
=>\(x=5-1\)
=>\(x=4\)
Vậy \(x=4\)
d,\(4x^3+15=47\)
=>\(4x^3=47-15\)
=>\(4x^3=32\)
=>\(x^3=32:4\)
=>\(x^3=8\)
=>\(x^3=2^3\)
=>\(x=2\)
Vậy\(x=2\)
e,\(\left(2x-1\right)^5=x^5\)
=>\(2x-1=x\)
=>\(2x-x=1\)
=>\(x=1\)
Vậy\(x=1\)
ĐÚNG K MÌNH NHA
\(a,\)( sửa lại xíu đề cho đúng nhé )
\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=-\frac{2x}{x^2+x+1}\)
\(\Rightarrow\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{3x^2}{\left(x-1\right)\left(x^2+x+1\right)}=-\frac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Rightarrow x^2+x+1-3x^2=-2x^2+2x\)
\(\Rightarrow x=1\)
\(g,\)\(\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)=-16\)
\(\Rightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)=-16\)
Đặt \(x^2+10x+16=a\)
\(\Rightarrow a\left(a+8\right)=-16\)
\(\Rightarrow a^2+8a+16=0\)
\(\Rightarrow\left(a+4\right)^2=0\)
\(\Leftrightarrow\left(x^2+10x+20\right)^2=0\)
\(\Rightarrow x^2+10x+25-25=0\)
\(\Rightarrow\left(x+5\right)^2-\left(\sqrt{5}\right)^2=0\)
\(\Rightarrow\left(x+5-\sqrt{5}\right)\left(x+5+\sqrt{5}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-5+\sqrt{5}\\x=-5-\sqrt{5}\end{cases}}\)
Bài 3:
a: Ta có: 60-3(x-2)=51
\(\Leftrightarrow x-2=3\)
hay x=5
b: Ta có: \(4x-20=25:2^2\)
\(\Leftrightarrow4x=\dfrac{25}{4}+20=\dfrac{105}{4}\)
hay \(x=\dfrac{105}{16}\)
c: Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)
\(\Leftrightarrow288:\left(x-3\right)^2=50-48=2\)
\(\Leftrightarrow\left(x-3\right)^2=144\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)
Bài1:
\(â,\left(x-4\right)^2-36=0\\ \Leftrightarrow\left(x-4\right)^2=36\\ \Leftrightarrow x-4\in\left\{-6;6\right\}\\ \Leftrightarrow x\in\left\{-2;10\right\}\)
Vậy...
b<Tương tự
c,\(x^2+8x+16=0\\ \Leftrightarrow\left(x+4\right)^2=0\\ \Leftrightarrow x+4=0\\ \Leftrightarrow x=-4\)
Vậy...
d,Tương tự
Bài2:
\(a,75^2-25^2\\ =\left(75-25\right)\left(75+25\right)\\ =100.50=5000\)
\(b,53^2-47^2\\ =\left(53-47\right)\left(53+47\right)\\ =6.100=600\)
Bài 1:
a) (x-4)^2 - 36 = 0
=> (x-4)^2 = 36
=> (x-4)^2 = 6^2
=> x-4 = 6
=>x = 2
b) (x-8)^2 = 121
=> (x-8)^2 = 11^2
=> x-8 = 11
=> x = 19
c) x^2 + 8x +16 = 0
=> x( x +8) = -16
=> x = -4
d) 4x^2 - 12x = -9( Mk chưa nghĩ ra !)
c) \(\dfrac{7x-1}{2}=5+\dfrac{9-5x}{6}\)
\(\Leftrightarrow\dfrac{6\left(7x-1\right)}{12}=\dfrac{5\cdot12}{12}+\dfrac{2\left(9-5x\right)}{12}\)
\(\Rightarrow42x-6=60+18-10x\)
\(\Leftrightarrow52x-84=0\)
\(\Leftrightarrow x=\dfrac{21}{13}\)
Vậy....
d) tương tự
a) \(\dfrac{x-3}{x-2}+\dfrac{x-2}{x-4}=-1\)ĐKXĐ : \(x\ne2;4\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\dfrac{\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=-1\)
\(\Leftrightarrow\dfrac{2x^2-11x+16}{x^2-6x+8}=-1\)
\(\Leftrightarrow2x^2-11x+16=-x^2+6x-8\)
\(\Leftrightarrow3x^2-17x+24=0\)
\(\Leftrightarrow3x^2-9x-8x+24=0\)
\(\Leftrightarrow3x\left(x-3\right)-8\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{8}{3}\end{matrix}\right.\)( thỏa mãn ĐKXĐ )
Vậy....
Đề bài đủ chưa? Yêu cầu chi nhỉ?