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a) \(\Leftrightarrow\left|2x-3\right|=\frac{1}{4}\Leftrightarrow\orbr{\begin{cases}x\ge\frac{3}{2}\mid:2x-3=\frac{1}{4}\Rightarrow2x=\frac{13}{4}\Rightarrow x=\frac{13}{8}\left(TM\right)\\x< \frac{3}{2}\mid:3-2x=\frac{1}{4}\Rightarrow2x=\frac{11}{4}\Rightarrow x=\frac{11}{8}\left(TM\right)\end{cases}.}\)
b) \(\Leftrightarrow\left|x-1\right|=\frac{3}{4}\Leftrightarrow\orbr{\begin{cases}x\ge1\mid:x-1=\frac{3}{4}\Rightarrow x=\frac{7}{4}\left(TM\right)\\x< 1\mid:1-x=\frac{3}{4}=>x=\frac{1}{4}\left(TM\right)\end{cases}}\)
c) \(\frac{3}{5\left(x-\frac{5}{6}\right)}-\frac{1}{2\left(\frac{3}{2}-1\right)}=-\frac{1}{4}\Leftrightarrow\frac{3}{\frac{5\left(6x-5\right)}{6}}-\frac{1}{2\cdot\frac{1}{2}}=-\frac{1}{4}\Leftrightarrow\frac{18}{5\left(6x-5\right)}=-\frac{1}{4}+1\)
\(\Leftrightarrow\frac{18}{5\left(6x-5\right)}=\frac{3}{4}\Leftrightarrow6x-5=\frac{24}{5}\Leftrightarrow6x=\frac{49}{5}\Leftrightarrow x=\frac{49}{30}\)
d) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\Leftrightarrow2\cdot\frac{x+1-2}{2\left(x+1\right)}=\frac{2015}{2016}\Leftrightarrow\frac{x-1}{x+1}=\frac{2015}{2016}\)
\(\Leftrightarrow2016x-2016=2015x+2015\Leftrightarrow x=2015+2016=4031\)
Vậy x = 4031.
Ta có : \(\left(2x+3\right)\left(x-7\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+3=0\\x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=-3\\x=7\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{3}{2}\\x=7\end{cases}}\)
Mình chỉ hướng dẫn giải thôi nhá chứ nhiều bài quá
a) Đặt \(\frac{x}{5}=\frac{y}{7}=k\Rightarrow x=5k;y=7k\)
Thay x.y=315 => 5k.7k=315 <=> 35k2=315 => k2=9 => k=3
x=5.3=15 ; y=7.3=21
b) 5x=9y<=> \(\frac{x}{9}=\frac{y}{5}\)
Theo TCDTSBN ta có : \(\frac{x}{9}=\frac{y}{5}=\frac{2x+3y}{2.9+3.5}=\frac{-33}{33}=-1\)
x/9=-1=>x=-9 ; y/5=-1=>y=-5
các bài còn lại tương tự b
B, => 2x+5=3x-8
2x-3x=-8-5
-x=-13
=>x=13
hoặc 2x+5=-3x+8
2x+3x=8-5
5x=3
x=\(\frac{3}{5}\)
\(\left|x+1\right|+\left|x-5\right|=7.\)
\(Th1:x+1< 0;x-5< 0\)
\(x< -1;x< 5\Rightarrow x< -1\)
\(\left|x+1\right|+\left|x-5\right|=7\)
\(-\left(x+1\right)-\left(x-5\right)=7\)
\(-x-1-x+5=7\)
\(-2x+4=7\)
\(-2x=3\)
\(x=-1,5\left(tm\right)\)
\(Th2:x+1>0;x-5>0\)
\(x>-1;x>5\Rightarrow x>5\)
\(\left|x+1\right|+\left|x-5\right|=7\)
\(x+1+x-5=7\)
\(2x-4=7\)
\(2x=11\)
\(x=5,5\)\(\left(tm\right)\)
\(Th3:x+1\le0;x-5>0\)
\(x\le-1;x>5\)(không xảy ra)
\(Th4:x+1>0;x-5\le0\)
\(x>-1;x\le5\Rightarrow-1< x\le5\)
\(\left|x+1\right|+\left|x-5\right|=7\)
\(-\left(x+1\right)+x-5=7\)
\(-x-1+x-5=7\)( không xảy ra)
Vậy x = -1,5 hoặc x = 5,5
\(\)
a) |x|=5
\(\Rightarrow\)x=5 hoặc x=-5
Vậy x=5 hoặc x=-5
b)|2x-1|=7
TH1: 2x-1=7
2x=7+1
2x=8
x=4
TH2:2x-1=-8
2x=-8+1
2x=-7
x=-7:2
x=-7/2
Vậy x= -7/2 hoặc x=4
a - x = 5 hoặc -5
b- =
th1 : 2x = 7+1
2x = 8
x= 8:2
x =4
th2 : 2x = -7 +1
2x = -6
x = -6 : 2
x = -3