\(\left\{{}\begin{matrix}xy=1100\\y-\dfrac{1100}{x+5}=2\end{matrix}\right.< =>\left\{{}\begin{matrix}y=\dfrac{1100}{x}\left(x\ne0\right)\left(1\right)\\\dfrac{1100}{x}-\dfrac{1100}{x+5}=2\left(2\right)\end{matrix}\right.\)

* giải pt(2)\(=>\dfrac{1100x+5500-1100x}{x^2+5x}=2\)

\(=>5500=2x^2+10x=>2x^2+10x-5500=0\)

\(=>\Delta=10^2-4\left(-5500\right)2=44100>0\)

\(=>\left[{}\begin{matrix}x1=\dfrac{-10+\sqrt{44100}}{2.2}=50\left(TM\right)\left(3\right)\\x2=\dfrac{-10-\sqrt{44100}}{2.2}=-55\left(TM\right)\left(4\right)\end{matrix}\right.\)

thế(3)(4) vào(1)\(=>\left[{}\begin{matrix}y=\dfrac{1100}{50}=22\\y=\dfrac{1100}{-55}=-20\end{matrix}\right.\)

vậy...

=>2x(x+5)=1100(x+5)-1100x

=>2x(x+5)=5500

=>2x^2+10x-5500=0

=>x=50 hoặc x=-55

đk : x khác 0 ; -5 

\(1100x+5500-1100x=2x\left(x+5\right)\)

\(\Leftrightarrow2x^2+10x-5500=0\Leftrightarrow x=50;x=-55\)(tm) 

= - 55 tm

a) Ta có :\(\dfrac{x+1}{111}=\dfrac{y+2}{222}=\dfrac{z+3}{333}=\dfrac{5x+5}{555}=\dfrac{2y+4}{444}\)

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

\(\dfrac{x+1}{111}=\dfrac{y+2}{222}=\dfrac{z+3}{333}=\dfrac{5x+5}{555}=\dfrac{2y+4}{444}\)\(=\dfrac{5x+2y+z}{555+444+333}=\dfrac{1100}{1332}=\dfrac{275}{333}\)

Từ đó tìm được x;y;z

b) Từ \(\dfrac{x}{2}=\dfrac{y}{3}\) \(\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{9}\)

Đặt \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=k\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=4k\\y^2=9k\end{matrix}\right.\)

\(\Rightarrow x^2\cdot y^2=4k\cdot9k=52\)

\(\Rightarrow36k^2=52\)

\(\Rightarrow k^2=\dfrac{13}{9}\) (sai đề)

b: Sửa đề: x^2+y^2=52

Đặt x/2=y/3=k

=>x=2k; y=3k

x^2+y^2=52

=>4k^2+9k^2=52

=>k^2=4

TH1: k=2

=>x=4; y=6

TH2: k=-2

=>x=-4; y=-6

c: Đặt x/5=y/3=k

=>x=5k; y=3k

x^2-y^2=16

=>25k^2-9k^2=16

=>k^2=1

TH1: k=1

=>x=5; y=3

TH2: k=-1

=>x=-5; y=-3

d: Đặt x/2=y/3=k

=>x=2k; y=3k

Ta có: xy=54

=>2k*3k=54

=>6k^2=54

=>k^2=9

TH1: k=3

=>x=6; y=9

TH2: k=-3

=>x=-6; y=-9

e: Đặt x/4=y/3=k

=>x=4k; y=3k

Ta có: xy=12

=>4k*3k=12

=>k^2=1

TH1: k=1

=>x=4; y=3

TH2: k=-1

=>x=-4; y=-3

\(A=\dfrac{3}{4}+\dfrac{3}{28}+\dfrac{3}{140}+...+\dfrac{3}{550}+\dfrac{3}{700}\)

\(A=\dfrac{3}{1.4}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{22.25}+\dfrac{3}{25.28}\)

\(A=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{22}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{28}\)

\(A=1-\dfrac{1}{28}\)

\(A=\dfrac{28}{28}-\dfrac{1}{28}=\dfrac{27}{28}\)

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

\(\dfrac{x+2}{x-2}-\dfrac{x-3}{x+2}=5\\ \Leftrightarrow\dfrac{\left(x+2\right)^2-\left(x-2\right)\left(x-3\right)}{\left(x-2\right)\left(x+2\right)}=5\\ \Leftrightarrow\dfrac{x^2+4x+4-x^2+5x-6}{\left(x^2-4\right)}=5\\ \Leftrightarrow9x-2=5\left(x^2-4\right)\\ \Leftrightarrow5x^2-20=9x-2\\ \Leftrightarrow5x^2-9x-18=0\\ \Leftrightarrow\left(5x+6\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x+6=0\\x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{6}{5}\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{6}{5};3\right\}\)

\(\dfrac{2}{x-14}-\dfrac{5}{x-13}=\dfrac{2}{x-9}-\dfrac{5}{x-11}\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne9;11;13;14\\\left(\dfrac{2}{x-14}-\dfrac{2}{3}\right)-\left(\dfrac{5}{x-13}-\dfrac{5}{4}\right)=\left(\dfrac{2}{x-9}-\dfrac{1}{4}\right)-\left(\dfrac{5}{x-11}-\dfrac{5}{6}\right)\end{matrix}\right.\)

\(\Leftrightarrow2\left(\dfrac{x-17}{3\left(x-14\right)}\right)-5\left(\dfrac{x-17}{4\left(x-13\right)}\right)=\left(\dfrac{x-17}{4\left(x-9\right)}\right)-5\left(\dfrac{x-17}{6\left(x-11\right)}\right)\)

\(\left(x-17\right)\left[\dfrac{2}{3\left(x-14\right)}-\dfrac{5}{4\left(x-13\right)}+\dfrac{5}{6\left(x-11\right)}-\dfrac{1}{4\left(x-9\right)}\right]=0\)

[..] vô nghiệm

x=17

Lời giải:

Bài của bạn ngonhuminh cơ bản không đúng do không có cơ sở khẳng định biểu thức trong ngoặc vuông vô nghiệm.

ĐKXĐ: \(x\neq \left\{9;11;13;14\right\}\)

\(\frac{2}{x-14}-\frac{5}{x-13}=\frac{2}{x-9}-\frac{5}{x-11}\)

\(\Leftrightarrow 2\left(\frac{1}{x-14}-\frac{1}{x-9}\right)=5\left(\frac{1}{x-13}-\frac{1}{x-11}\right)\)

\(\Leftrightarrow \frac{10}{(x-14)(x-9)}=\frac{10}{(x-13)(x-11)}\)

\(\Rightarrow (x-14)(x-9)=(x-13)(x-11)\)

\(\Leftrightarrow x^2-23x+126=x^2-24x+143\)

\(\Leftrightarrow x-17=0\Leftrightarrow x=17\)

Thử lại thấy thỏa mãn.

Vậy \(x=17\)

không vô nghiệm thì tìm đi

ĐKXĐ: ...

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

\(\Leftrightarrow\dfrac{5+x^2}{x^2}+\dfrac{2x}{\sqrt{5+x^2}}=3\)

Đặt \(\dfrac{x}{\sqrt{5+x^2}}=t\)

\(\Rightarrow\dfrac{1}{t^2}+2t=3\)

\(\Rightarrow2t^3-3t^2+1=0\)

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

\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{\sqrt{5+x^2}}=1\left(x>0\right)\\\dfrac{x}{\sqrt{5+x^2}}=-\dfrac{1}{2}\left(x< 0\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{x^2}{5+x^2}=1\left(vn\right)\\\dfrac{x^2}{5+x^2}=\dfrac{1}{4}\left(x< 0\right)\end{matrix}\right.\)

\(\Rightarrow x=-\sqrt{\dfrac{5}{3}}\)