\(\dfrac{x-3}{7}=\dfrac{2x-7}{13}\)

\(7\left(2x-7\right)=13\left(x-3\right)\)

\(14x-49=13x-39\)

\(14x-13x=49-39\)

\(x=10\)

Bạn cần giúp nhanh nhưng lại không ghi đầy đủ đề bài?

ĐKXĐ: \(x\ne y,x\ne-y\)

\(hpt\Leftrightarrow\left(\dfrac{1}{x+y}+\dfrac{1}{x-y}\right)-\left(\dfrac{1}{x+y}+\dfrac{1}{x-y}\right)=\dfrac{5}{8}-\dfrac{3}{8}\)

\(\Leftrightarrow0=\dfrac{1}{4}\left(VLý\right)\)

Vậy hpt vô nghiệm

má bài này lol thắng cx đăng tr :vv

\(a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac{1}{z}\Rightarrow\left\{{}\begin{matrix}a+b+c=2\\2ab-c^2=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=2-a-b\\2ab-\left(2-a-b\right)^2=4\end{matrix}\right.\Leftrightarrow}}\left\{{}\begin{matrix}c=2-a-b\\2ab-4-a^2-b^2+4a+4a-2ab-4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=2-a-b\\\left(a-2\right)^2+\left(b-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=2\\c=-2\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{1}{2}\\z=-\dfrac{1}{2}\end{matrix}\right.\)

a) Ta có: \(\left(2x+7\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow\left(2x+7\right)^2-\left(x+3\right)^2=0\)

\(\Leftrightarrow\left(2x+7-x-3\right)\left(2x+7+x+3\right)=0\)

\(\Leftrightarrow\left(x+4\right)\cdot\left(3x+10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\3x+10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\3x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-\dfrac{10}{3}\end{matrix}\right.\)

Vậy: \(S=\left\{-4;-\dfrac{10}{3}\right\}\)

b) Ta có: \(\left(4x+14\right)^2=\left(7x+2\right)^2\)

\(\Leftrightarrow\left(4x+14\right)^2-\left(7x+2\right)^2=0\)

\(\Leftrightarrow\left(4x+14-7x-2\right)\left(4x+14+7x+2\right)=0\)

\(\Leftrightarrow\left(-3x+12\right)\left(11x+16\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-3x+12=0\\11x+16=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=-12\\11x=-16\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{16}{11}\end{matrix}\right.\)Vậy: \(S=\left\{4;-\dfrac{16}{11}\right\}\)

(2x+7)²=(x+3)²

=>(2x+7)²-(x+3)²=0

=>(2x+7-x-3)(2x+7+x+3)=0

=>(x-4)(3x+10)=0

=>x-4=0 hoặc 3x+10=0

TH1:x-4=0=>x=4

TH2:3x+10=0=>x=-10/3

(4x+14)²=(7x+2)²

(4x+14)²-(7x+2)²=0

(4x+14-7x-2)(4x+14+7x+2)=0

(-3x+12)(11x+16)=0

TH1:-3x+12=0=>x=4

TH2:11x+16=0=>x=-16/11

Đề bài là gì bạn nhỉ ?

a: =>4x^2-4x+1+7>4x^2+3x+1

=>-4x+8>3x+1

=>-7x>-7

=>x<1

b: \(\Leftrightarrow12x+1>=36x+12-24x-3\)

=>1>=9(loại)

a) \(\dfrac{x-3}{x+5}=\dfrac{5}{7}\)

⇔\(7\left(x-3\right)=5\left(x+5\right)\)

⇔\(7x-21=5x+25\)

⇔\(7x-21-5x-25=0\)

⇔\(2x-46=0\)

⇔\(2x=46\)

⇔\(x=23\)

b) \(\dfrac{7}{x-1}=\dfrac{x+1}{9}\)

⇔\(\left(x+1\right)\left(x-1\right)=7.9\)

⇔\(x^2-1=63\)

⇔\(x^2=64=8^2\)

⇔\(\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

\(a.\dfrac{x-3}{x+5}=\dfrac{5}{7}\)                          

    \(\left(x-3\right).7=\left(x+5\right).5\) 

    \(\left(x.7\right)+\left[\left(-3\right).7\right]=\left(x.5\right)+\left(5.5\right)\)

    \(7x-21=5x+25\)

    \(7x-5x=25+21\)

             \(2x=46\)

               \(x=46:2\)

               \(x=23\)

Câu b/ cứ làm theo câu a/ là được

AcCl₃ và NiCl₂

\(\Rightarrow\dfrac{a}{b}\times\left(-\dfrac{2}{7}+\dfrac{5}{7}\right)=\dfrac{5}{7}\\ \Rightarrow\dfrac{a}{b}\times\dfrac{3}{7}=\dfrac{5}{7}\\ \Rightarrow\dfrac{a}{b}=\dfrac{5}{7}:\dfrac{3}{7}\\ \Rightarrow\dfrac{a}{b}=\dfrac{5}{3}\\ \Rightarrow a=5;b=3\)

\(=>\dfrac{a}{b}\times\left(-\dfrac{2}{7}+\dfrac{5}{7}\right)=\dfrac{5}{7}\)

\(=>\dfrac{a}{b}\times\dfrac{3}{7}=\dfrac{5}{7}=>\dfrac{a}{b}=\dfrac{5}{7}:\dfrac{3}{7}=\dfrac{5}{3}\)

vậy \(\dfrac{a}{b}=\dfrac{5}{3}\)

Đặt : \(\dfrac{x}{5}=\dfrac{y}{3}=k\)

`=>x=5k,y=3k`

Ta có : \(x^2-y^2=4=>\left(5k\right)^2-\left(3k\right)^2=4\\ =>25k^2-9k^2=4\\ =>16k^2=4\\ =>k^2=\dfrac{1}{4}\\ =>k=\pm\dfrac{1}{2}\)

\(=>\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{5}{2}\\y=\dfrac{3}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=-\dfrac{3}{2}\end{matrix}\right.\end{matrix}\right.\)

4*cos(pi/6-a)*sin(pi/3-a)

=4*(cospi/6*cosa+sinpi/6*sina)*(sinpi/3*cosa-sina*cospi/3)

=4*(căn 3/2*cosa+1/2*sina)*(căn 3/2*cosa-1/2*sina)

=4*(3/4*cos^2a-1/4*sin^2a)

=3cos^2a-sin^2a

=3(1-sin^2a)-sin^2a

=3-4sin^2a

=>m=3; n=-4

m^2-n^2=-7