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\(\left|2+3x\right|=\left|4x-3\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2+3x=4x-3\\2+3x=3-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\x=\frac{1}{7}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{7};5\right\}\)
\(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)
Vậy \(x\in\left\{\frac{1}{11};\frac{3}{5}\right\}\)
\(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)
\(\Leftrightarrow\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)
Giải tiếp tương tự
Sau đó giải tiếp câu còn lại
a) \(\left(4x^2-2\right)^2=\frac{196}{81}\)
<=> \(2^2\left(2x^2-1\right)^2=\frac{196}{81}\)
<=> \(4\left(2x^2-1\right)^2=\frac{196}{81}\)
<=> \(\left(2x^2-1\right)^2=\frac{196}{81}:4\)
<=> \(\left(2x^2-1\right)^2=\frac{49}{81}\)
<=> \(2x^2-1=\pm\sqrt{\frac{49}{81}}\)
<=> \(2x^2-1=\pm\frac{7}{9}\)
<=> \(\orbr{\begin{cases}2x^2-1=\frac{7}{9}\\2x^2-1=-\frac{7}{9}\end{cases}}\)<=> \(\orbr{\begin{cases}x=\pm\frac{2\sqrt{2}}{3}\\x=\pm\frac{1}{3}\end{cases}}\)
=> \(\orbr{\begin{cases}x=\pm\frac{2\sqrt{2}}{3}\\x=\pm\frac{1}{3}\end{cases}}\)
Ta có : lũy thừa mũ chẵn luôn lớn hơn hoặc bằng 0 và GTTĐ luôn lớn hơn hoặc bằng 0, mà theo đề bài
=> +) 4x - 7 = 0
4x = 7
x = 7/4
=> +) 7 - 4x = 0
7 = 4x
x = 7/4
=> x = 7/4
Vậy,............
a) \(0,25\left(x+\frac{1}{2}\right)+\frac{3}{4}+x=\frac{1}{2}\)
\(\Leftrightarrow0,25x+\frac{1}{8}+\frac{3}{4}+x=\frac{1}{2}\)
\(\Leftrightarrow1,25x=-\frac{3}{8}\)
\(\Leftrightarrow x=-\frac{3}{10}\)
c) \(2x^2+4x=0\)
\(\Leftrightarrow2x\left(x+2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x+2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=-2\end{array}\right.\)
a) 0,25(x+1/2) + 3/4 + x= 1/2
<=> \(0,25x+\frac{1}{8}+\frac{3}{4}+x=\frac{1}{2}\)
<=> \(\frac{5}{4}x=\frac{1}{2}-\frac{1}{8}-\frac{3}{4}=-\frac{3}{8}\)
<=> x=\(-\frac{3}{10}\)
B) 1/2 ÷(x+7/5)-1/5=0.75
<=> \(\frac{1}{2}:\left(x+\frac{7}{5}\right)-\frac{1}{5}=\frac{3}{4}\)
<=> \(\frac{1}{2x}+\frac{5}{14}-\frac{1}{5}=\frac{3}{4}\)
<=> \(\frac{1}{2x}=\frac{3}{4}+\frac{1}{5}-\frac{5}{14}=\frac{83}{140}\)
<=> x=\(\frac{70}{83}\)
C) 2x^2 + 4x= 0
\(x\left(x+2\right)=0\)
<=> x=0 hoặc x=-2
D) x^2 + 4x = 0<=> x(x+4)=0
<=> x=0 hoặc x=-4
\(\left[-\frac{5}{4}x+2,15\right]\left[2\frac{3}{7}-(-\frac{1}{2}x)\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}-\frac{5}{4}x+2,15=0\\2\frac{3}{7}-\left[-\frac{1}{2}x\right]=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{43}{25}\\x=-\frac{34}{7}\end{cases}}\)
\(\Leftrightarrow\left|4x-7\right|\left(\left|4x-7\right|-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left|4x-7\right|=0\\\left|4x-7\right|=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x-7=0\\4x-7=5\\7-4x=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}\\x=3\\x=\dfrac{1}{2}\end{matrix}\right.\)