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AT
12 tháng 6 2017
K chép lại đề, lm luôn nhé:
*\(\Rightarrow\) \(\left(\dfrac{7}{2}+2x\right)\cdot\dfrac{8}{3}=\dfrac{16}{3}\)
\(\Rightarrow\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)
\(\Rightarrow2x=2-\dfrac{7}{2}=-\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{3}{4}\)
* \(\Rightarrow\left|2x-\dfrac{2}{3}\right|=\dfrac{\dfrac{3}{4}-2}{2}=-\dfrac{5}{8}\)
=> K có gt x nào t/m đề
* Đề sai
* \(\Rightarrow\left[{}\begin{matrix}3x-1=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
*\(\Rightarrow\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=-\dfrac{21}{4}\)
\(\Rightarrow2x-1=\dfrac{1}{3}:\left(-\dfrac{21}{4}\right)=-\dfrac{4}{63}\)
\(\Rightarrow2x=-\dfrac{4}{63}+1=\dfrac{59}{63}\)
\(\Rightarrow x=\dfrac{59}{63}:2=\dfrac{59}{126}\)
* \(\Rightarrow\left(2x+\dfrac{3}{5}\right)^2=\dfrac{9}{25}\)
\(\Rightarrow\left[{}\begin{matrix}2x+\dfrac{3}{5}=\dfrac{3}{5}\\2x+\dfrac{3}{5}=-\dfrac{3}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=0\Rightarrow x=0\\2x=-\dfrac{6}{5}\Rightarrow x=-\dfrac{3}{5}\end{matrix}\right.\)
* \(\Rightarrow-5x-1-\dfrac{1}{2}x+\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(\Rightarrow-5x-\dfrac{1}{2}x-\dfrac{3}{2}x=-\dfrac{5}{6}+1-\dfrac{1}{3}\)
\(\Rightarrow-7x=-\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{1}{6}:7=-\dfrac{1}{42}\)
12 tháng 6 2017
a)\(\left(3\dfrac{1}{2}+2x\right).2\dfrac{2}{3}=5\dfrac{1}{3}\)
\(\left(\dfrac{7}{2}+2x\right).\dfrac{8}{3}=\dfrac{16}{3}\)
\(\dfrac{7}{2}+2x=\dfrac{16}{3}:\dfrac{8}{3}=2\)
\(2x=2-\dfrac{7}{2}=\dfrac{-3}{2}\Rightarrow x=\dfrac{-3}{4}\)
b)\(\dfrac{3}{4}-2.\left|2x-\dfrac{2}{3}\right|=2\)
\(2.\left|2x-\dfrac{2}{3}\right|=\dfrac{3}{4}-2=\dfrac{-1}{4}\)
\(\Rightarrow\left|2x-3\right|=\dfrac{-1}{8}\)
\(\Rightarrow x\in\varnothing\)
c) Đề sai,bạn có viết chữ x đâu,đó là phép tính mà.
d)\(\left(3x-1\right)\left(\dfrac{-1}{2}x+5\right)=0\)
\(\Leftrightarrow3x-1=0\Rightarrow x=\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{-1}{2}x+5=0\Rightarrow x=10\)
e)\(\dfrac{1}{4}+\dfrac{1}{3}:\left(2x-1\right)=-5\)
\(\dfrac{1}{3}:\left(2x-1\right)=-5-\dfrac{1}{4}=\dfrac{-21}{4}\)
\(2x-1=\dfrac{1}{3}:\dfrac{-21}{4}=\dfrac{-4}{63}\)
\(\Rightarrow2x=\dfrac{59}{63}\Rightarrow x=\dfrac{59}{126}\)
g)\(\left(2x+\dfrac{3}{5}\right)^2-\dfrac{9}{25}=0\)
\(\left(2x+\dfrac{3}{5}\right)^2=0+\dfrac{9}{25}=\dfrac{9}{25}\)
\(\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2=\left(\dfrac{-3}{5}\right)^2\)
\(th1:x=0\)
\(th2:x=\dfrac{-3}{5}\)
h)\(-5\left(x+\dfrac{1}{5}\right)-\dfrac{1}{2}\left(x-\dfrac{2}{3}\right)=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(-5x+-1-\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{3}{2}x-\dfrac{5}{6}\)
\(\Leftrightarrow-5x+-1+\dfrac{5}{6}-\dfrac{1}{3}=2x\)
\(-5x+\dfrac{-1}{2}=2x\)
\(\dfrac{-1}{2}=2x+5x\)
\(\dfrac{-1}{2}=7x\Rightarrow x=\dfrac{-1}{14}\)
KL
3 tháng 8 2017
1. a, 3x + |x - 2| = 8
<=> |x - 2| = 8 - 3x
Xét 2 TH :
TH1: x - 2 = 8 - 3x
<=> x + 3x = 8 + 2
<=> 4x = 10
<=> x = \(\dfrac{5}{2}\) (thỏa mãn)
TH2: x - 2 = -(8 - 3x)
<=> x - 2 = -8 + 3x
<=> -2 + 8 = 3x - x
<=> 6 = 2x
<=> x = 3 (thỏa mãn)
b, 5 - |x - 1| = 4
<=> |x - 1| = 1
<=> \(\left[{}\begin{matrix}x-1=1\\x-1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\) (thỏa mãn)
@Nguyễn Hoàng Vũ
KL
3 tháng 8 2017
2. 5.(x - 2) - 4.(1 - 3x) = |3 - 7| + 2.(1 + 2x)
<=> 5x - 10 - 4 + 12x = 4 + 2 + 4x
<=> 17x - 14 = 6 + 4x
<=> 17x - 4x = 6 + 14
<=> 13x = 20
<=> x = \(\dfrac{20}{13}\) (thỏa mãn)
@Nguyễn Hoàng Vũ
DQ
11 tháng 4 2019
\(\left(2x-3\right)\left(6-2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1,5\\x=3\end{matrix}\right.\)
DQ
11 tháng 4 2019
*\(\frac{1}{4}+\frac{1}{3}:\left(2x-1\right)=-5\)
\(\Leftrightarrow\frac{1}{3}\cdot\frac{1}{2x-1}=-5-\frac{1}{4}\)
\(\Leftrightarrow\frac{1}{3\left(2x-1\right)}=\frac{-21}{4}\)
\(\Leftrightarrow-63\left(2x-1\right)=4\)
\(\Leftrightarrow2x-1=-\frac{4}{63}\)
\(\Leftrightarrow2x=\frac{59}{63}\)
\(x=\frac{59}{126}\)
18 tháng 7 2017
a ) Ta có : 4(x - 5) - 3(x + 7) = -19
<=> 4x - 20 - 3x - 21 = -19
=> x - 41 = -19
=> x = -19 + 41
=> x = 22
b) Ta có " 7(x - 3) - 5(3 - x) = 11x - 5
<=> 7x - 21 - 15 + 5x = 11x - 5
<=> 12x - 36 = 11x - 5
=> 12x - 11x = -5 + 36
=> x = 31
BA
4 tháng 10 2019
\(\left(2x+1\right)^3=125\)
\(\left(2x+1\right)^3=5^3\)
\(2x+1=5\)
\(2x=4\)
\(x=2\)
\(b,x^6=x^2\)
\(x^6-x^2=0\)
\(x^2\cdot\left(x^4-1\right)=0\)
\(\orbr{\begin{cases}x^2=0\\x^4-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
\(c\text{}\text{}\text{}\text{},\left(x-2\right)\cdot\left(x-5\right)=0\)
\(\orbr{\begin{cases}x-2=0\\x-5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=5\end{cases}}}\)
\(d,x^{10}-x^5=0\)
\(x^5\cdot\left(x^5-1\right)=0\)
\(\orbr{\begin{cases}x^5=0\\x^5=1\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)
\(e,\left(x-5\right)^4=\left(x-5\right)^6\)
\(\left(x-5\right)^4-\left(x-5\right)^6=0\)
\(\left(x-5\right)^4\cdot\left[1-\left(x-5\right)^2\right]=0\)
\(\orbr{\begin{cases}\left(x-5\right)^4=0\\1-\left(x-5\right)^2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\pm1+5\end{cases}}}\)
\(\hept{\begin{cases}x=5\\x=6\\x=4\end{cases}}\)
VC
Vũ Cao Minh
CTV
VIP
4 tháng 10 2019
\(\left(2x+1\right)^3=125\Rightarrow\left(2x+1\right)^3==5^3\Rightarrow2x+1=5\)
\(\Rightarrow2x=5-1=4\Rightarrow x=4:2=2\)
\(x^6=x^2\Rightarrow x^2.x^4=x^2\)Vì vậy nên \(x=\pm1\)
\(\left(x-2\right)\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\Rightarrow x=0+2=5\\x-5=0\Rightarrow X=0+5=5\end{cases}}\)
\(\left(x-1\right)\left(3-x\right)+4=4\)
\(\Rightarrow\left(x-1\right)\left(3-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\3-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}}\)
vậy_
\(5-\left(x+1\right)\cdot2x=5\)
\(\Rightarrow\left(x+1\right)\cdot2x=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\2x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}}\)
vậy_
(x-1).(3-x) + 4 = 4
(x-1).(3-x) = 0
=> x - 1 = 0=> x = 1
3-x = 0 => x = 3
KL:...
5-(x+1).2x = 5
(x+1).2x = 0
=> x + 1 = 0 => x = -1
2x = 0 => x = 0
KL:...