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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) 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
\(1,\dfrac{2x+4}{7}=\dfrac{4x-2}{15}=\dfrac{2.\left(2x+4\right)}{2.7}=\dfrac{4x+8}{14}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{2x+4}{7}=\dfrac{4x-2}{15}==\dfrac{4x+8}{14}=\dfrac{\left(4x+8\right)-\left(4x-2\right)}{14-15}=\dfrac{10}{-1}=-10\)
\(\Rightarrow\dfrac{2x+4}{7}=-10\)
\(\Rightarrow2x+4=-10.7=-70\)
\(\Rightarrow2x=-70+4=-66\)
\(\Rightarrow x=-66:2=-33\)
Vậy \(x=-33\)
\(2,\dfrac{2x+3}{5}=\dfrac{7x-3}{15}=\dfrac{7.\left(2x+3\right)}{7.5}=\dfrac{2.\left(7x-3\right)}{2.15}=\dfrac{14x+21}{35}=\dfrac{14x-6}{30}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\dfrac{2x+3}{5}=\dfrac{14x+21}{35}=\dfrac{14x-6}{30}=\dfrac{\left(14x+21\right)-\left(14x-6\right)}{35-30}=\dfrac{29}{5}\)
\(\Rightarrow\dfrac{2x+3}{5}=\dfrac{29}{5}\)
\(\Rightarrow2x+3=29\)
\(\Rightarrow2x=29-3=26\)
\(\Rightarrow x=26:2=13\)
\(3,\dfrac{11x-2}{7x+5}=\dfrac{11}{8}\)
\(\Rightarrow\dfrac{11x-2}{11}=\dfrac{7x+5}{8}=\dfrac{7.\left(11x-2\right)}{7.11}=\dfrac{11.\left(7x+5\right)}{8.11}=\dfrac{77x-14}{77}=\dfrac{77x+55}{88}=\dfrac{\left(77x+55\right)-\left(77x-14\right)}{88-77}=\dfrac{69}{11}\)
\(\Rightarrow\dfrac{11x-2}{11}=\dfrac{69}{11}\)
\(\Rightarrow11x-2=69\)
\(\Rightarrow11x=69+2=71\)
\(\Rightarrow x=\dfrac{71}{11}\)
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
b: 2x-3<0
=>2x<3
hay x<3/2
c: \(\left(2x-4\right)\left(9-3x\right)>0\)
=>(x-2)(x-3)<0
=>2<x<3
d: \(\dfrac{2}{3}x-\dfrac{3}{4}>0\)
=>2/3x>3/4
hay x>9/8
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. )
\(a,\dfrac{2}{3}-\dfrac{1}{3}\left(x-\dfrac{3}{2}\right)-\dfrac{1}{2}\left(2x+1\right)=5\)
\(\dfrac{2}{3}-\dfrac{1}{3}x-\dfrac{1}{2}-x+\dfrac{1}{2}=5\)
\(\dfrac{2}{3}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}x-x=5\)
\(\dfrac{2}{3}-\dfrac{1}{3}x-x=5\)
\(\dfrac{2}{3}-\dfrac{4}{3}x=5\)
\(\dfrac{4}{3}x=\dfrac{2}{3}-5\)
\(\dfrac{4}{3}x=-\dfrac{13}{3}\)
\(x=-\dfrac{13}{3}:\dfrac{4}{3}\)
\(x=-\dfrac{13}{4}\)
Vậy...............
\(b,\left(x+\dfrac{1}{2}\right)\left(\dfrac{3}{4}-x\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{3}{4}-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy................
\(c,\dfrac{2x-1}{-3+2}=0\)
\(\Rightarrow2x-1=0\)
\(\Rightarrow x=\dfrac{1}{2}\)
Vậy.............
a) \(\dfrac{1,2}{x+3}=\dfrac{5}{4}\)
\(\Rightarrow\left(x+3\right).5=1,2.4\)
\(\Rightarrow\left(x+3\right).5=4,8\)
\(\Rightarrow x+3=4,8:5\)
\(\Rightarrow x+3=0,96\)
\(\Rightarrow x=-2,04\)
vậy \(x=-2,04\)
b)\(\dfrac{3}{5}:\dfrac{2x}{15}=\dfrac{1}{2}:\dfrac{4}{5}\)
\(\Rightarrow\dfrac{3}{5}:\dfrac{2x}{15}=\dfrac{5}{8}\)
\(\Rightarrow\dfrac{2x}{15}=\dfrac{3}{5}:\dfrac{5}{8}\)
\(\Rightarrow\dfrac{2x}{15}=\dfrac{24}{25}\)
\(\Rightarrow15.24=\left(2x\right).25\)
\(\Rightarrow360=\left(2x\right).25\)
\(\Rightarrow360:25=2x\)
\(\Rightarrow14,4=2x\)
\(\Rightarrow x=7,2\)
vậy \(x=7,2\)
\(a,\dfrac{1,2}{x+3}=\dfrac{5}{4}\\ \left(x+3\right).5=1,2.4\\ 5x+8=4,8\\ 5x=4,8-8\\ 5x=-3,2\\ x=-3,2:5=-0,64\)
\(b,\dfrac{3}{5}:\dfrac{2x}{15}=\dfrac{1}{2}:\dfrac{4}{5}\\ \dfrac{2x}{15}=\dfrac{3}{5}\cdot\dfrac{4}{5}:\dfrac{1}{2}\\ \dfrac{2x}{15}=\dfrac{12}{25}.2\\ \dfrac{2x}{25}=\dfrac{24}{25}\\ 2x=\dfrac{24}{25}.5\\ 2x=\dfrac{24}{5}\\ x=\dfrac{24}{5}\cdot\dfrac{1}{2}=\dfrac{12}{5}\)
\(c,-\dfrac{4}{2,5}:3,5=1,5:x\\ x=3,5.1,5:\left(-\dfrac{4}{25}\right)\\ x=\dfrac{21}{4}\cdot\left(-\dfrac{25}{4}\right)=-\dfrac{525}{16}\)
\(d,0,12:3=2x:\dfrac{3}{5}\\ 2x=0,12\cdot\dfrac{3}{5}:3\\ 2x=\dfrac{9}{125}\cdot\dfrac{1}{3}\\ 2x=\dfrac{3}{125}\\ x=\dfrac{3}{125}\cdot\dfrac{1}{2}=\dfrac{3}{250}\)
\(\dfrac{x-2}{4}=\dfrac{y+1}{5}=\dfrac{z+3}{7}\)
\(\Rightarrow\dfrac{2\left(x-2\right)}{8}=\dfrac{y+1}{5}=\dfrac{2\left(z+3\right)}{14}\)
\(\Rightarrow\dfrac{2x-4}{8}=\dfrac{y+1}{5}=\dfrac{2z+6}{14}\)
Dựa vào tính chất dãy tỉ số bằng nhau ta có:
\(=\dfrac{2x-4+y+1-2z-6}{8+5-14}\)
\(=\dfrac{2x+y-2z-9}{-1}\)
\(=\dfrac{7-9}{-1}=2\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-2}{4}=2\Rightarrow x-2=8\Rightarrow x=10\\\dfrac{y+1}{5}=2\Rightarrow y+1=10\Rightarrow y=9\\\dfrac{z+3}{7}=2\Rightarrow z+3=14\Rightarrow z=11\end{matrix}\right.\)
\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{3}=\dfrac{z}{5}\)
\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12};\dfrac{y}{12}=\dfrac{z}{20}\)
\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)
\(\Rightarrow\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}\)
\(=\dfrac{2x-3y+z}{18-36+20}\)
\(=\dfrac{6}{2}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.9=27\\y=3.12=36\\z=3.20=60\end{matrix}\right.\)
\(\dfrac{2x}{3}=\dfrac{3y}{4}=\dfrac{4z}{5}\)
\(\Rightarrow x.\dfrac{2}{3}=y.\dfrac{3}{4}=z.\dfrac{4}{5}\)
\(\Rightarrow x:\dfrac{3}{2}=y:\dfrac{4}{3}=z:\dfrac{5}{4}\)
\(\Rightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)
\(=\dfrac{x+y+z}{\dfrac{3}{2}+\dfrac{4}{3}+\dfrac{5}{4}}\)
\(=\dfrac{49}{\dfrac{49}{12}}=12\)
\(\Rightarrow\left\{{}\begin{matrix}x=12.\dfrac{3}{2}=18\\y=12.\dfrac{4}{3}=16\\z=12.\dfrac{5}{4}=15\end{matrix}\right.\)
Ta có :
\(\dfrac{x}{3}=\dfrac{y}{4}=>\dfrac{x}{9}=\dfrac{y}{12}\left(1\right)\)
\(\dfrac{y}{3}=\dfrac{z}{5}=>\dfrac{y}{12}=\dfrac{z}{20}\left(2\right)\)
Từ (1),(2)=>\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)=\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{2x-3y+z}{18-36+20}=\dfrac{6}{2}=3\)
=>\(\left\{{}\begin{matrix}x=27\\y=36\\z=60\end{matrix}\right.\)
Lời giải:
PT $\Leftrightarrow \frac{2x-1}{203}+1)+(\frac{2x-3}{205}+1)=(\frac{5-2x}{207}-1)-(\frac{2x}{101}+2)+5$
$\Leftrightarrow \frac{2x+202}{203}+\frac{2x+202}{205}=\frac{-(2x+202)}{207}-\frac{2x+202}{101}+5$
$\Leftrightarrow (2x+202)(\frac{1}{203}+\frac{1}{205}+\frac{1}{207}+\frac{1}{101})=5$
$\Leftrightarrow x=\frac{1}{2}[5: (\frac{1}{203}+\frac{1}{205}+\frac{1}{207}+\frac{1}{101})-202]$