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a) Ta có \(|5\left(2x+3\right)\ge0\)
\(|2\left(2x+3\right)|\ge0\)
\(|2x+3|\ge0\)
\(\Rightarrow|5\left(2x+3\right)|+|\left(2x+3\right)|+|2x+3|\ge0\)
\(\Rightarrow5\left(2x+3\right)+2\left(2x+3\right)+2x+3=16\)
\(\Rightarrow10x+15+4x+6+2x+3=16\)
\(\Rightarrow\left(10x+4x+2x\right)+\left(15+6+3\right)=16\)
\(\Rightarrow16x+24=16\)
\(\Rightarrow24=16x-16\)
\(\Rightarrow24=x\)
Vậy x=24
1) \(\left(\frac{2x}{3}-3\right):\left(-10\right)=\frac{2}{5}\)
\(\Leftrightarrow-\frac{\frac{2x}{3}-3}{10}=\frac{2}{5}\)
\(\Leftrightarrow-\left(\frac{\frac{2x}{3}}{10}-\frac{3}{10}\right)=\frac{2}{5}\)
\(\Leftrightarrow-\left(\frac{2x}{3\times10}-\frac{3}{10}\right)=\frac{2}{5}\)
\(\Leftrightarrow-\left(\frac{2x}{30}-\frac{3}{10}\right)=\frac{2}{5}\)
\(\Leftrightarrow-\frac{x}{15}+\frac{3}{10}=\frac{2}{5}\)
\(\Leftrightarrow\frac{3}{10}-\frac{x}{15}=\frac{2}{5}\)
\(\Leftrightarrow-\frac{x}{15}=\frac{2}{5}-\frac{3}{10}\)
\(\Leftrightarrow-\frac{x}{15}=\frac{1}{10}\)
\(\Leftrightarrow-x=\frac{15}{10}\)
\(\Leftrightarrow-x=\frac{3}{2}\)
\(\Leftrightarrow x=-\frac{3}{2}\)
Vậy \(x=-\frac{3}{2}\)
2) \(\left|2x-1\right|+1=4\)
\(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=4\\2x=-2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-1\end{cases}}}\)
Vậy \(x\in\left\{2;-1\right\}\)
Ta có: 10x=6y=5z⇔x110=y16=z1510x=6y=5z⇔x110=y16=z15 và x+y−z=24x+y−z=24
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
x110=y16=z15=x+y−z110+16−15=24:115=360x110=y16=z15=x+y−z110+16−15=24:115=360
=> x = 360 : 10 = 36
y = 360 : 6 = 60
z = 360 : 5 = 72
dựa theo lm nhá ! chúc học tốt
#)Giải :
a) \(\left(5x+1\right)^2=\frac{36}{49}\Leftrightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\Leftrightarrow5x+1=\frac{6}{7}\Leftrightarrow5x=-\frac{1}{7}\Leftrightarrow x=-\frac{1}{35}\)
b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\Leftrightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2=\frac{4}{9}\Leftrightarrow x=\frac{2}{3}\)
c) \(\left(8x-1\right)^{2x+1}=5^{2x+1}\Leftrightarrow8x-1=5\Leftrightarrow8x=6\Leftrightarrow x=\frac{6}{8}\)
a) \(\left(5x+1\right)^2=\frac{36}{49}\)
\(\left(5x+1\right)^2=\frac{6^2}{7^2}\)
\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)
\(\Leftrightarrow5x+1=\frac{6}{7}\)
\(5x=\frac{6}{7}-1\)
\(5x=\frac{6}{7}-\frac{7}{7}\)
\(5x=-\frac{1}{7}\)
\(x=-\frac{1}{7}\div5\)
\(x=-\frac{1}{7}\times\frac{1}{5}\)
\(x=-\frac{1}{35}\)
Vậy \(x=-\frac{1}{35}\)
a) [2x] = -1\(\Rightarrow-1\le2x< 0\Rightarrow-0,5\le x< 0\)
b) [x + 0,4] = 3\(\Rightarrow3\le x+0,4< 4\Rightarrow2,6\le x< 3,6\)
c)\(\left[\frac{2}{3}x-5\right]=3\Rightarrow3\le\frac{2}{3}x-5< 4\Rightarrow8\le\frac{2}{3}x< 9\Rightarrow12\le x< 13,5\)
Từ bài trên,ta có :\(\left[x\right]=y\Rightarrow y\le x< y+1\left(x\in Q;y\in Z\right)\)
\(\left(2x-3\right)^2=36\)
\(\Leftrightarrow\left(2x-3\right)^2=6^2\)
\(\Leftrightarrow2x-3=6\)
\(\Leftrightarrow2x=9\)
\(\Leftrightarrow x=\frac{9}{2}\)
Tìm x biết:
(2x−3)2= 36
=>(2x-3)2=62
=>2x-3=6
2x=9
x=9:2
x=4,5
Vậy x=4,5