Ta có: \(3\times y\times\left(\frac11\times\frac12+\frac12\times\frac13+\cdots+\frac15\times\frac16\right)=\frac34\)

=>\(3\times y\times\left(1-\frac12+\frac12-\frac13+\cdots+\frac15-\frac16\right)=\frac34\)

=>\(3\times y\times\left(1-\frac16\right)=\frac34\)

=>\(y\times3\times\frac56=\frac34\)

=>\(y\times\frac52=\frac34\)

=>\(y=\frac34:\frac52=\frac34\times\frac25=\frac{6}{20}=\frac{3}{10}\)

Về học lại hằng đẳng thức nha .-.

\(\Leftrightarrow\dfrac{\left(x+2\right)+5}{2-x}=\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}\\ \Leftrightarrow-\left(x+2\right)+5\left(x+2\right)=2x-3\\ \Leftrightarrow6x+12-2x+3=0\\ \Leftrightarrow4x+15=0\\ \Leftrightarrow x=\dfrac{-15}{4}\)

\(\dfrac{1}{x+2}+\dfrac{5}{2-x}=\dfrac{2x-3}{x^2-4}\)

\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{5}{x-2}-\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}=0\left(đk:x\ne\pm2\right)\)

\(\Leftrightarrow\dfrac{x-2-5\left(x+2\right)-2x-3}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow x-2-5x-10-2x-3=0\)

\(\Leftrightarrow-6x-15=0\)

\(\Leftrightarrow-6x=15\)

\(\Leftrightarrow x=-\dfrac{15}{6}\left(n\right)\)

Vậy \(S=\left\{-\dfrac{15}{6}\right\}\)

\(\dfrac{1}{x+2}+\dfrac{5}{2-x}=\dfrac{2x-3}{x^2-4}\) đkxđ : x khác 2 , x khác -2.

<=> \(\dfrac{1}{x+2}-\dfrac{5}{x-2}-\dfrac{2x-3}{x^2-4}=0\)

<=> \(\dfrac{1.\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{5.\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}=0\)

<=> \(\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}-\dfrac{2x-3}{\left(x-2\right)\left(x+2\right)}=0\)

<=>\(x-2-5x-10-2x+3=0\)

<=> \(-6x-9=0\)

<=> \(x=-\dfrac{9}{6}=-\dfrac{3}{2}\left(nhận\right)\)

Vậy pt có nghiệm \(S=\left\{-\dfrac{3}{2}\right\}\)

a: Bạn ghi lại đề nha bạn

b: \(30\left(x+2\right)-6\left(x-5\right)-24x=100\)

=>\(30x+60-6x+30-24x=100\)

=>\(\left(30x-6x-24x\right)+\left(60+30\right)=100\)

=>0x=100-90=10(vô lý)

c: \(\left(x-7\right)\left(x+3\right)< 0\)

TH1: \(\left\{{}\begin{matrix}x-7>0\\x+3< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>7\\x< -3\end{matrix}\right.\)

=>\(x\in\varnothing\)

TH2: \(\left\{{}\begin{matrix}x-7< 0\\x+3>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\)

=>-3<x<7

mà x nguyên

nên \(x\in\left\{-2;-1;0;1;2;3;4;5;6\right\}\)

d: -1<2x-1<4

=>\(-1+1< 2x< 4+1\)

=>0<2x<5

=>0<x<2,5

mà x nguyên

nên \(x\in\left\{1;2\right\}\)

thank you friend nhiều

\(\left(x-1\right)^2=\left(x-1\right)^4\)

\(\Rightarrow\left(x-1\right)^4-\left(x-1\right)^2=0\)

\(\Rightarrow\left(x-1\right)^2\left[\left(x-1\right)^2-1\right]=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\\left(x-1\right)^2=1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x-1=1\\x-1=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=2\\x=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(\dfrac{x-7}{y-6}=\dfrac{7}{6}\)

⇔ \(\dfrac{x-7}{7}=\dfrac{y-6}{6}\)

⇔ \(\dfrac{x}{7}=\dfrac{y}{6}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{7}=\dfrac{y}{6}=\dfrac{x-y}{7-6}=-4\)

⇒ \(\left\{{}\begin{matrix}x=-28\\y=-24\end{matrix}\right.\)

Vậy ...

bn chưa học nhưng ngta học r:D

\(\dfrac{x+1}{x-2}=\dfrac{1}{x^2-4}ĐK:x\ne\pm2\)

\(\Rightarrow\left(x+1\right)\left(x+2\right)=1\Leftrightarrow x^2+3x+2=1\)

\(\Leftrightarrow x^2+3x+1=0\)

=> Phương trình vô nghiệm 

thật ra bài này vẫn có nghiệm nhưng nghiệm là số vô tỉ 

\(\Leftrightarrow x^2+3x+1=0\Leftrightarrow x^2+3x+\dfrac{9}{4}-\dfrac{5}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{3}{2}\right)^2-\left(\dfrac{\sqrt{5}}{2}\right)^2=0\)nhưng lớp 8 mình chưa làm nên mình để pt vô nghiệm nhé 

a: \(\left|3x-2\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)

b: Ta có: \(\left|5x-3\right|=\left|x-7\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-3=x-7\\5x-3=7-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-4\\6x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{3}\end{matrix}\right.\)

Theo đề ta có : x + 1 chia hết cho 2, 4, 5  và x là số nhỏ nhất hay x + 1 thuộc BCNN(2, 4, 5)

Ta có: 2 = 2 ; 4 = 2²  ; 5 = 5

=> BCNN(2, 4, 5) = 2² . 5 = 20

=> x + 1 = 20 => x = 20 - 1= 19

Vậy x = 19 

x chia 2 dư 1; x chia 4 dư 3; x chia 5 dư 4

\(\Rightarrow x+1\in BC\left(2,4,5\right)=B\left(20\right)=\left\{20;40;...\right\}\)

Mà \(x\) nhỏ nhất nên \(x-1\) nhỏ nhất

\(\Rightarrow x+1=20\Rightarrow x=19\)

\(\dfrac{x}{2}=\dfrac{y}{4};\dfrac{y}{5}=\dfrac{z}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{20}=\dfrac{z}{12}\)

Áp dụng t/c của dãy số bằng nhau, ta có: \(\dfrac{x-y+z}{10-20+12}=\dfrac{4}{2}=2\)

\(\dfrac{x}{10}=2\Rightarrow x=20\)

\(\dfrac{y}{20}=2\Rightarrow y=40\)

\(\dfrac{z}{12}=2\Rightarrow z=24\)

x/10=y/20=z/12

x-y+z/=10-20+12=4/2=2

x=2.10=20

y=2.20=40

z=2.12=24