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\(\frac{2x+3}{7}=\frac{4x-1}{15}\)
=> \(\frac{15\left(2x+3\right)}{105}=\frac{7\left(4x-1\right)}{105}\)
=> \(15\left(2x+3\right)=7\left(4x-1\right)\)
=> \(30x+45=28x-7\)
=>\(28x-30x=45+7\)
=> \(-2x=52\)
=>\(x=52:-2\)
=> \(x=-26\)
a . \(\dfrac{x+1}{x-2}=\dfrac{3}{4}\)
=> \(\left(x+1\right).4=3.\left(x-2\right)\)
=> \(4x+4=3x-6\)
=> \(4x-3x=-6-4\)
=> x = -10
b. \(\dfrac{2x-3}{x+1}=\dfrac{4}{7}\)
=> \(\left(2x-3\right).7=4.\left(x+1\right)\)
=> \(14x-21=4x+4\)
=> \(14x-4x=4+21\)
=> \(10x=25\)
=> \(x=\dfrac{5}{2}\)
c. \(\dfrac{2x+4}{7}=\dfrac{4x-2}{15}\)
=> \(\left(2x+4\right).15=\left(4x-2\right).7\)
=> \(30x+60=28x-14\)
=> \(30x-28x=-14-60\)
=> \(2x=-74\)
=> \(x=-37\)
#Yiin
a, \(\dfrac{x+1}{x-2}=\dfrac{3}{4}\Rightarrow4\left(x+1\right)=3\left(x-2\right)\)
\(\Rightarrow4x+4=3x-6\)
\(\Rightarrow4x-3x=-6-4\Rightarrow x=-10\)
b, \(\dfrac{2x-3}{x+1}=\dfrac{4}{7}\Rightarrow7\left(2x-3\right)=4\left(x+1\right)\)
\(\Rightarrow14x-21=4x+4\)
\(\Rightarrow14x-4x=4+21\Rightarrow10x=25\Rightarrow x=\dfrac{5}{2}\)
c, \(\dfrac{2x+4}{7}=\dfrac{4x-2}{15}\Rightarrow15\left(2x+4\right)=7\left(4x-2\right)\)
\(\Rightarrow30x+60=28x-14\)
\(\Rightarrow30x-28x=-14-60\)
\(\Rightarrow2x=-74\Rightarrow x=-37\)
\(\frac{x-1}{-15}=\frac{-60}{x-1}\)
\(\Leftrightarrow\left(x-1\right)^2=900\\ \Leftrightarrow\left(x-1\right)^2=\left(\pm30\right)^2\\ \Rightarrow x-1\in\left\{30;-30\right\}\)
\(\Rightarrow\left[{}\begin{matrix}x-1=30\\x-1=-30\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=31\\x=-29\end{matrix}\right.\)
Vậy...
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
Mấy bài dễ tự làm nhé:D
1)
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)
Ta có điều phải chứng minh
\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)
Ta có điều phải chứng minh
\(\text{a) }3-2\left|4x-5\right|=\dfrac{2}{6}\\ \Leftrightarrow2\left|4x-5\right|=\dfrac{8}{3}\\ \Leftrightarrow\left|4x-5\right|=\dfrac{4}{3}\\ \Leftrightarrow4x-5=-\dfrac{4}{3}\text{ hoặc :}\\ 4x-5=-\dfrac{4}{3}\\ \Leftrightarrow\left[{}\begin{matrix}4x-5=-\dfrac{4}{3}\\4x-5=\dfrac{4}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{11}{3}\\4x=\dfrac{19}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{12}\\x=\dfrac{19}{12}\end{matrix}\right.\\ \text{Vậy }x=\dfrac{11}{12}\text{ hoặc }x=\dfrac{19}{12}\)
a. Áp dụng t/c dãy tỉ sô bằng nhau ta có :
\(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{3y}{12}=\dfrac{x-3y}{3-12}=\dfrac{36}{-9}=-4\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-4\Rightarrow x=-12\\\dfrac{y}{4}=-4\Rightarrow y=-16\end{matrix}\right.\)
Vậy.............
b. Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x+3y}{4+9}=\dfrac{39}{13}=3\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=3\Rightarrow x=6\\\dfrac{y}{3}=3\Rightarrow y=9\end{matrix}\right.\)
Vậy.........
c. Áp dụng t/c dãy tỉ số bằng nhau ta có :
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{4x}{12}=\dfrac{3y}{15}=\dfrac{4x-3y}{12-15}=\dfrac{12}{-3}=-4\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{3}=-4\Rightarrow x=-12\\\dfrac{y}{5}=-4\Rightarrow y=-20\end{matrix}\right.\)
Vậy............
a, \(\dfrac{x}{3}=\dfrac{y}{4}\Rightarrow\dfrac{x}{3}=\dfrac{3y}{12}\)
Áp dụng t/c dãy tỉ số = nhau ,ta có :
\(\dfrac{x}{3}=\dfrac{3y}{12}=\dfrac{x-3y}{3-12}=\dfrac{36}{-9}=-4\)
\(\Rightarrow\dfrac{x}{3}=\dfrac{y}{4}=-4\Rightarrow\left\{{}\begin{matrix}x=-12\\y=-16\end{matrix}\right.\)
Vậy ...
b,c tương tự
a,|x2−13x2−13| = 3232
b, 32−1232−12 ( 2x-1)=3434
c, |x-1|+2x=2
a)\(\left|\dfrac{x}{2}-\dfrac{1}{3}\right|=\dfrac{3}{2}\)
TH1
\(\dfrac{x}{2}-\dfrac{1}{3}=\dfrac{3}{2}\)
=>\(\dfrac{x}{2}=\dfrac{11}{6}\)
=>x=\(\dfrac{11.2}{6}\)
=>x=\(\dfrac{11}{3}\)
TH2
\(\dfrac{x}{2}-\dfrac{1}{2}=-\dfrac{3}{2}\)
=>\(\dfrac{x}{2}=-\dfrac{3}{2}+\dfrac{1}{2}\)
=>\(\dfrac{x}{2}=-1\)
=>x=-2
\(\frac{2x+3}{7}=\frac{4x-1}{15}\\ \Rightarrow15\left(2x+3\right)=7\left(4x-1\right)\\ \Rightarrow30x+45=28x-7\\ \Rightarrow2x=-52\\ \Rightarrow x=-26\)