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b: =>(3x-1)(3x+1)(2x+3)=0
hay \(x\in\left\{\dfrac{1}{3};-\dfrac{1}{3};-\dfrac{3}{2}\right\}\)
c: \(\Leftrightarrow\left|2x-\dfrac{1}{3}\right|=\dfrac{5}{6}+\dfrac{3}{4}=\dfrac{19}{12}\)
=>2x-1/3=19/12 hoặc 2x-1/3=-19/12
=>2x=23/12 hoặc 2x=-15/12=-5/4
=>x=23/24 hoặc x=-5/8
d: \(\Leftrightarrow-\dfrac{5}{6}\cdot x+\dfrac{3}{4}=-\dfrac{3}{4}\)
=>-5/6x=-3/2
=>x=3/2:5/6=3/2*6/5=18/10=9/5
e: =>2/5x-1/2=3/4 hoặc 2/5x-1/2=-3/4
=>2/5x=5/4 hoặc 2/5x=-1/4
=>x=5/4:2/5=25/8 hoặc x=-1/4:2/5=-1/4*5/2=-5/8
f: =>14x-21=9x+6
=>5x=27
=>x=27/5
h: =>(2/3)^2x+1=(2/3)^27
=>2x+1=27
=>x=13
i: =>5^3x*(2+5^2)=3375
=>5^3x=125
=>3x=3
=>x=1
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
a: \(\Leftrightarrow\dfrac{5}{3}+\dfrac{4}{3}< x< 3+\dfrac{1}{5}+1+\dfrac{4}{5}\)
=>3<x<5
=>x=4
b: \(\Leftrightarrow\dfrac{1}{3}:2x=-5+\dfrac{1}{4}=-\dfrac{19}{4}\)
=>\(2x=\dfrac{1}{3}:\dfrac{-19}{4}=\dfrac{1}{3}\cdot\dfrac{-4}{19}=\dfrac{-4}{57}\)
=>x=-2/57
c: \(\Leftrightarrow x\cdot\dfrac{-3}{2}=\dfrac{10}{3}-\dfrac{6}{7}=\dfrac{70-18}{21}=\dfrac{52}{21}\)
=>\(x=\dfrac{-52}{21}:\dfrac{3}{2}=\dfrac{-52}{21}\cdot\dfrac{2}{3}=\dfrac{-104}{63}\)
d: \(\Leftrightarrow70+18< x< 120+70\)
=>88<x<190
hay \(x\in\left\{89;90;...;188;189\right\}\)
a) Ta có: |2x-5| \(\ge\)0 với mọi x
mà |2x-5|=-4
=> x\(\in\varnothing\)
b)\(\dfrac{1}{3}-\left|\dfrac{5}{4}-2x\right|=\dfrac{1}{4}\)
=>\(\left|\dfrac{5}{4}-2x\right|=\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{1}{12}\)
=>\(\left[{}\begin{matrix}\dfrac{5}{4}-2x=\dfrac{1}{12}\\\dfrac{5}{4}-2x=-\dfrac{1}{12}\end{matrix}\right.=>\left[{}\begin{matrix}2x=\dfrac{5}{4}-\dfrac{1}{12}=\dfrac{7}{6}\\2x=\dfrac{5}{4}+\dfrac{1}{12}=\dfrac{4}{3}\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=\dfrac{7}{12}\\x=\dfrac{2}{3}\end{matrix}\right.\)
phần c và d cũng tương tự bạn tự làm nha
a,\(x+\frac{1}{4}=\frac{3}{4} \)
<=>x=\(\frac{3}{4}-\frac{1}{4} \)
<=>\(x=\frac{1}{2} \)
b,\(\frac{3}{4}-\frac{2}{5}x=\frac{29}{60} \)
<=>\(\frac{2}{5}x =\frac{3}{4}-\frac{29}{60} \)
<=>\(\frac{2}{5}x=\frac{4}{15} \)
<=x=\(\frac{2}{3} \)
c,\(2x-\frac{1}{3}=\frac{-5}{6} \)
2x=\(\frac{-5}{6}+\frac{1}{3} \)
2x=\(\frac{-1}{2} \)
x=\(\frac{-1}{4} \)
d,2-\(\frac{3}{4x}=\frac{1}{2} \)
<=>\(\frac{3}{4x}=2-\frac{1}{2} \)
<=>\(\frac{3}{4}x=\frac{3}{2} \)
<=>x=2
e,\(\frac{11}{12}- \frac{2}{3}|x| =\frac{3}{8} \)
<=>\(\frac{2}{3}|x|=\frac{13}{24} \)
<=>\(|x|=\frac{13}{16} \)
<=>x=\(\pm\frac{13}{16} \)
f,\(|2x-1|=5\)
<=>2x-1=5 hoặc 2x-1=-5
<=> 2x=6 2x=-4
<=> x=3 x=-2
Giải
a) \(x+\dfrac{1}{4}=\dfrac{3}{4}\)
=> \(x=\dfrac{3}{4}-\dfrac{1}{4}\)=>\(x=\dfrac{1}{2}\)
b)\(\dfrac{3}{4}-\dfrac{2}{5}x=\dfrac{29}{60}\)
=>\(\dfrac{2}{5}x=\dfrac{3}{4}-\dfrac{29}{60}\)=>\(\dfrac{2}{5}x=\dfrac{4}{15}\)
=>\(x=\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{2}{3}\)
c)\(2x-\dfrac{1}{3}=\dfrac{-5}{6}\)=>\(2x=\dfrac{-5}{6}+\dfrac{1}{3}\)=\(\dfrac{-1}{2}\)
=>\(x=\dfrac{-1}{2}:2=\dfrac{-1}{4}\)
d)\(2-x:\dfrac{3}{4}=\dfrac{1}{2}\)=>\(x:\dfrac{3}{4}=2-\dfrac{1}{2}\)=\(\dfrac{3}{2}\)
=>\(x=\dfrac{3}{2}.\dfrac{3}{4}=\dfrac{9}{8}\)
e)\(\dfrac{11}{12}-\dfrac{2}{3}.\left|x\right|=\dfrac{3}{8}\)=>\(\dfrac{2}{3}.\left|x\right|=\dfrac{11}{12}-\dfrac{3}{8}\)=\(\dfrac{13}{24}\)
=>\(\left|x\right|=\dfrac{13}{24}:\dfrac{2}{3}=\dfrac{13}{16}\)
Vậy: \(x=\dfrac{13}{16}\)hoặc\(x=\dfrac{-13}{16}\)
f)\(\left|2x-1\right|=5\)
*2x-1=5 =>2x=5+1=6 =>x=6:2=3
*2x-1=-5 =>2x=(-5)+1=-4 =>x=-4:2=-2
Vậy: x=3 hoặc x=-2
Tick cho Phong nhé:>
Yêu nhiều>3
#Phong_419
câu E
\(\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left(2x-5\right)\left(5-2x\right)=-\left(\dfrac{3}{2}\right)^4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{5}{2}\\\left|2x-5\right|=\left(\dfrac{3}{2}\right)^2\end{matrix}\right.\)
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x< \dfrac{5}{2}\\2x-5=-\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{11}{8}< \dfrac{5}{2}\left(n\right)\end{matrix}\right.\\\left\{{}\begin{matrix}x>\dfrac{5}{2}\\2x-5=\left(\dfrac{3}{2}\right)^2\Rightarrow x=\dfrac{29}{8}>\dfrac{5}{2}\left(n\right)\end{matrix}\right.\end{matrix}\right.\)
câu F (bạn cho vào lớp 7.2=lớp 14 nhé. )
a: \(\left|x\right|=3+\dfrac{1}{5}=\dfrac{16}{5}\)
mà x<0
nên x=-16/5
b: \(\left|x\right|=-2.1\)
nên \(x\in\varnothing\)
c: \(\left|x-3.5\right|=5\)
=>x-3,5=5 hoặc x-3,5=-5
=>x=8,5 hoặc x=-1,5
d: \(\left|x+\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
=>|x+3/4|=1/2
=>x+3/4=1/2 hoặc x+3/4=-1/2
=>x=-1/4 hoặc x=-5/4
VÍ DỤ
Giải:
• Xét đẳng thức véc tơ: 1X1 2X2 3X3 X
• Suy ra hệ phương trình có ma trận mở rộng nhận X1, X2, X3, X làm các cột
2 3 1 1 2 3 1 1 2