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Bài giải
a, \(\left(x-5\cdot9\right)\cdot2=0\)
\(x-45=0\)
\(x=45\)
b, \(40\cdot\left(x-2\right)=40\)
\(x-2=1\)
\(x=3\)
c, \(\left(x-4\right)\cdot3+2=101\)
\(\left(x-4\right)\cdot3=99\)
\(x-4=33\)
\(x=37\)
Bài làm :
\(a,\left(x-5.9\right).2=0\)
\(\Rightarrow x-45=0\)
\(\Rightarrow x=0+45\)
\(\Rightarrow x=45\)
\(b,40.\left(x-2\right)=40\)
\(\Rightarrow x-2=40:40\)
\(\Rightarrow x-2=1\)
\(\Rightarrow x=3\)
\(c,\left(x-4\right).3+2=101\)
\(\Rightarrow\left(x-4\right).3=99\)
\(\Rightarrow x-4=99:3\)
\(\Rightarrow x-4=33\)
\(\Rightarrow x=37\)
Học tốt nhé
Bài giải
a) (2.x - 1).(x - 3) = 0 (x thuộc N)
Mà 0.0 = 0 hoặc 0 nhân với số nào cũng bằng 0
Suy ra một trong hai biểu thức "(2.x - 1)" hoặc "x - 3" = 0
Ta có:
2.x - 1 = 0 | x - 3 = 0 |
2.x = 0 + 1 | x = 0 + 3 |
2.x = 1 | x = 3 (chọn) |
x = 1 : 2 | |
x = \(\frac{1}{2}\)(loại vì x thuộc N) |
Vậy x = 3
Mấy câu còn lại để mai mình làm
a) Ta có:
(2x - 1)(x - 3) = 0
=> \(\orbr{\begin{cases}2x-1=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\left(ktm\right)\\x=3\left(tm\right)\end{cases}}\)
b) 3x(x - 2) = x - 2
=> 3x(x - 2) - (x - 2) = 0
=> (3x - 1)(x - 2) = 0
=> \(\orbr{\begin{cases}3x-1=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{3}\left(tkm\right)\\x=2\left(tm\right)\end{cases}}\)
c) (2x + 3)x - 2(2x + 3) = 0
=> (2x + 3)(x - 2) = 0
=> \(\orbr{\begin{cases}2x+3=0\\x-2=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-\frac{3}{2}\left(ktm\right)\\x=2\left(tm\right)\end{cases}}\)
d) 4(x - 3) + 2x(x - 3) = 0
=> (4 + 2x)(x - 3) = 0
=> \(\orbr{\begin{cases}4+2x=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-2\left(tkm\right)\\x=3\left(tm\right)\end{cases}}\)
bài 4 tìm X biết
a)(X-36):18=12
X-36= 12*18
X-36= 216
X= 216 + 36
X= 252
b)5X-3-2 . 52 =52 . 3
5X-3 = 52 . 3 + 2 . 52
5X-3 = 52 . ( 3+2)
5X-3 = 52 . 5
5X-3 = 52 . 5\(^1\)
5X-3 = 5\(^3\)
=> X-3=3
X= 3+3
X=6
c)X . (X - 3)=0
\(\rightarrow\) \(\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\)
\(\rightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
D)3X+1-32=2 .32
3X+1 = 2 .32 + 32
3X+1 = 2 .32 + 32 .1
3X+1 = 32 . ( 2+1)
3X+1 = 32 .3
3X+1 = 32 .3\(^1\)
3X+1 = 3\(^3\)
=> X+1= 3
X= 3+1
X= 4
a) \(\frac{3}{2}-\left(x-\frac{7}{3}\right)=\left|-\frac{3}{4}-\frac{9}{8}\right|\)
=> \(\frac{3}{2}-x+\frac{7}{3}=\left|-\frac{15}{8}\right|\)
=> \(\frac{3}{2}-x+\frac{7}{3}=\frac{15}{8}\)
=> \(\frac{3}{2}-x=-\frac{11}{24}\)
=> \(x=\frac{47}{24}\)
b) \(\frac{5}{2}-\left(\frac{3}{2}-\frac{7}{3}+x\right)=\frac{8}{15}-\left(\frac{1}{4}-\frac{7}{10}\right)\)
=> \(\frac{5}{2}-\frac{3}{2}+\frac{7}{3}-x=\frac{8}{15}-\left(-\frac{9}{20}\right)\)
=> \(\frac{10}{3}-x=\frac{59}{60}\)
=> \(x=\frac{10}{3}-\frac{59}{60}=\frac{47}{20}\)
c) \(2\left(\frac{3}{4}-5x\right)=\frac{4}{5}-3x\)
=> \(\frac{3}{2}-10x-\frac{4}{5}+3x=0\)
=> \(\left(\frac{3}{2}-\frac{4}{5}\right)+\left(-10x+3x\right)=0\)
=> \(\frac{7}{10}-7x=0\)
=> \(7x=\frac{7}{10}\)
=> x = 1/10
\(\frac{x-2}{2}=\frac{18}{x-2}\)
\(\Rightarrow\left(x-2\right)^2=18\times2\)
\(\Leftrightarrow\left(x-2\right)^2=36\)
\(\Leftrightarrow\left(x-2\right)^2=6^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=-6\\x-2=6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=8\end{cases}}\)
Vậy \(x\in\left\{-4;8\right\}\)
\(\frac{x-2}{2}=\frac{18}{x-2}\)
\(\Rightarrow\left(x-2\right).\left(x-2\right)=18.2\)
\(\Rightarrow\left(x-2\right)^2=36\)
\(\Rightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=8\\x=-4\end{cases}}\)
Chúc bạn học tốt !!!
Ta có:\(\hept{\begin{cases}\left(x-2011\right)^2\ge0\forall x\\|y+2012|\ge0\forall y\end{cases}}\)
\(\Rightarrow\left(x-2011\right)^2+|y+2012|\ge0\forall x,y\)
Do đó \(\left(x-2011\right)^2+|y+2012|=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x-2011\right)^2=0\\|y+2012|=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x-2011=0\\y+2012=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x=2011\\y=-2012\end{cases}}\)
Vậy ...
\(\left(x-2011\right)^2+\left|y+2012\right|=0\)
Ta có: \(\hept{\begin{cases}\left(x-2011\right)^2\ge0\\\left|y+2012\right|\ge0\end{cases}}\Rightarrow\)\(\left(x-2011\right)^2+\left|y+2012\right|\ge0\)
Dấu "="\(\Leftrightarrow\hept{\begin{cases}\left(x-2011\right)^2\ge0\\\left|y+2012\right|\ge0\end{cases}}\))
Vậy \(\left(x-2011\right)^2+\left|y+2012\right|=0\Leftrightarrow\hept{\begin{cases}x=2011\\y=-2012\end{cases}}\)
\(\frac{x}{5}-\frac{3}{7}=\frac{10}{13}\)
\(\Rightarrow\frac{91x}{455}-\frac{195}{455}=\frac{350}{455}\)
\(\Rightarrow91x-195=350\)
\(\Rightarrow91x=545\)
\(\Rightarrow x=\frac{545}{91}\)
\(\frac{-3x}{4}+\frac{1}{3}=\frac{11}{5}\)
\(\Rightarrow\frac{-45x}{60}+\frac{20}{60}=\frac{132}{60}\)
\(\Rightarrow-45x+20=132\)
\(\Rightarrow-45x=112\)
\(\Rightarrow x=-\frac{112}{45}\)
`@` `\text {Ans}`
`\downarrow`
`[ (100 - x)*2 - 2] = 40`
`=> (100-x)*2 - 2 = 40`
`=> (100 - x)*2 = 40 + 2`
`=> (100 - x)*2 = 42`
`=> 100 - x = 42 \div 2`
`=> 100 - x = 21`
`=> x = 100 - 21`
`=> x = 79`
Vậy, `x=79.`
\(\left[\left(100-x\right).2-2\right]=40\)
\(=>\left(100-x\right).2=40+2\)
\(=>\left(100-x\right).2=42\)
\(=>100-x=42:2\)
\(=>100-x=21\)
\(=>x=100-21\)
\(=>x=79\)
Chúc bạn học tốt