Bạn ơi đề có sai ko

Sao lại \(\dfrac{y}{y}\)

Mik xin loi, de dung la

\(\dfrac{x}{3}=\dfrac{y}{4};\dfrac{y}{y}=\dfrac{z}{8}\)va \(3x-2y-z=13\)

a: \(\dfrac{x+1}{5}+\dfrac{x+1}{6}=\dfrac{x+1}{7}+\dfrac{x+1}{8}\)

\(\Leftrightarrow\left(x+1\right)\left(\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}-\dfrac{1}{8}\right)=0\)

=>x+1=0

hay x=-1

b: \(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)

=>x-2010=0

hay x=2010

c: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

\(\Leftrightarrow\dfrac{x}{\left(x+2\right)\left(x+17\right)}=\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}\)

=>x=15

a,\(x-\dfrac{3}{5}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}+\dfrac{3}{5}\)

\(x=\dfrac{6}{5}\)

b,\(\left|x\right|-\dfrac{4}{5}=\dfrac{2}{5}\)

\(\left|x\right|=\dfrac{2}{5}+\dfrac{4}{5}\)

\(\left|x\right|=\dfrac{6}{5}\)

\(\Rightarrow x=\pm\dfrac{6}{5}\)

c,\(\dfrac{x}{-5}=\dfrac{24}{15}\)

\(x=\dfrac{-5.24}{15}\)

\(x=\dfrac{-24}{5}\)

d,Áp dụng tc dãy TSBN, ta có:

\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x-y}{4-5}=\dfrac{21}{-1}=-21\)

+\(\dfrac{x}{4}=-21\Rightarrow x=-21.4=-84\)

+\(\dfrac{y}{5}=-21\Rightarrow y=-21.5=-105\)

Vậy x=-84 ; y=-105

a/ \(x-\dfrac{3}{5}=\dfrac{3}{5}\)

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

\(\Leftrightarrow x=\dfrac{6}{5}\)

Vậy...

b/ \(\left|x\right|-\dfrac{4}{5}=\dfrac{2}{5}\)

\(\Leftrightarrow\left|x\right|=\dfrac{2}{5}+\dfrac{4}{5}\)

\(\Leftrightarrow\left|x\right|=\dfrac{6}{5}\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\\x=-\dfrac{6}{5}\end{matrix}\right.\)

Vậy...

c/ \(\dfrac{x}{-5}=\dfrac{24}{15}\)

\(\Leftrightarrow15x=-120\)

\(\Leftrightarrow x=-8\)

Vậy...

c/ Theo t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{4}=\dfrac{y}{5}=\dfrac{x-y}{4-5}=\dfrac{21}{-1}=-21\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{4}=-21\\\dfrac{y}{5}=-21\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-84\\y=-105\end{matrix}\right.\)

Vậy..

ê

2x-3y+5z=1 hoặc =-1

TH1: \(\dfrac{x}{y}\)=\(\dfrac{3}{2}\)=>\(\dfrac{x}{3}\)=\(\dfrac{y}{2}\)=>\(\dfrac{x}{15}\)=\(\dfrac{y}{10}\)

\(\dfrac{y}{z}\)=\(\dfrac{5}{7}\)=>\(\dfrac{y}{5}\)=\(\dfrac{z}{7}\)=>\(\dfrac{y}{10}\)=\(\dfrac{z}{14}\)

\(\Rightarrow\)\(\dfrac{x}{15}\)=\(\dfrac{y}{10}\)=\(\dfrac{z}{14}\)=>\(\dfrac{2x}{30}\)=\(\dfrac{3y}{30}\)=\(\dfrac{5z}{70}\)

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

\(\dfrac{2x-3y+5z}{30-30+70}\)=\(\dfrac{1}{70}\)

=>x=1.15:7=\(\dfrac{3}{14}\)

y=\(\dfrac{1}{7}\)

z=\(\dfrac{1}{5}\)

TH2:............=-1 tự tính nhé làm tương tựmình còn phải ôn bài

1.a)\(2.x-\dfrac{5}{4}=\dfrac{20}{15}\)

\(\Leftrightarrow2.x=\dfrac{20}{15}+\dfrac{5}{4}=\dfrac{4}{3}+\dfrac{5}{4}=\dfrac{16+15}{12}=\dfrac{31}{12}\)

\(\Leftrightarrow x=\dfrac{31}{12}:2=\dfrac{31}{12}.\dfrac{1}{2}=\dfrac{31}{24}\)

b)\(\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{8}\right)\)

\(\Leftrightarrow\left(x+\dfrac{1}{3}\right)^3=\left(-\dfrac{1}{2}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{3}=-\dfrac{1}{2}\)

\(\Leftrightarrow x=-\dfrac{1}{2}-\dfrac{1}{3}=-\dfrac{5}{6}\)

2.Theo đề bài, ta có: \(\dfrac{a}{2}=\dfrac{b}{3}\) và \(a+b=-15\)

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

\(\dfrac{a}{2}=\dfrac{b}{3}=\dfrac{a+b}{2+3}=\dfrac{-15}{5}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{2}=-3\Rightarrow a=-6\\\dfrac{b}{3}=-3\Rightarrow b=-9\end{matrix}\right.\)

3.Ta xét từng trường hợp:

-TH1:\(\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\)\(\Rightarrow x\in\left\{0;1\right\}\)

-TH2:\(\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\)\(\Rightarrow x\in\varnothing\)

Vậy \(x\in\left\{0;1\right\}\)

4.\(B=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{9}{49}\right)^9=\left(\dfrac{3}{7}\right)^{21}:\left[\left(\dfrac{3}{7}\right)^2\right]^9=\left(\dfrac{3}{7}\right)^{21}:\left(\dfrac{3}{7}\right)^{18}=\left(\dfrac{3}{7}\right)^3=\dfrac{27}{343}\)

Câu 1: Mình chỉnh sửa lại đầu bài của bạn nha. Không biết có đúng không. Nếu để đầu bài như bạn thì mình không làm ra được. Mog góp ý !!!!

Áp dụng t/c DTSBN ta có:

\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=x+y+z\)

\(=\dfrac{x+y+x}{y+z+1+x+z+1+x+y-2}=\dfrac{x+y+x}{2x+2y+2z}=\dfrac{1}{2}\)

=>\(\dfrac{x}{y+z+1}=\dfrac{1}{2}\left(1\right)\)

=>\(\dfrac{y}{x+z+1}=\dfrac{1}{2}\left(2\right)\)

=>\(\dfrac{z}{x+y-2}=\dfrac{1}{2}\left(3\right)\)

=> x+y+z = 1/2 (4)

Ta có : Từ (1) => 2x = y+z+1 kết hợp (4)

=> 2x = 1/2-x+1

=> 3x = 3/2 => x=1/2

Ta có: Từ (2) => 2y = x+z+1

=> 2y + y = x+y+z+1

=> 3y = 1/2+1 (theo 4) => 3y=3/2

=> y=1/2

Ta có : Từ (4) => x+y+z=1/2

=>1/2 + 1/2 +z = 1/2

=> z=-1/2

Vậy ( x;y;z)=(1/2;1/2;-1/2)

a) Giải

Vì \(5x=2y=3z\)

\(\Rightarrow\dfrac{5x}{30}=\dfrac{2y}{30}=\dfrac{3z}{30}\)

\(\Rightarrow\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{10}\)

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

\(\dfrac{x}{6}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{x+y-z}{6+15-10}=\dfrac{33}{11}=3\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{6}=3\Rightarrow x=18\\\dfrac{y}{15}=3\Rightarrow y=45\\\dfrac{z}{10}=3\Rightarrow z=30\end{matrix}\right.\)

Vậy \(x=18,\) \(y=45\) hoặc \(z=30.\)

c) Giải

(Vì mk bt bạn bấm nhầm nên đề bị sai, mk sửa 7 \(\rightarrow\) y do trên bàn phím, 7 với y ở vị trí gần nhau mà 2 với y ở cách xa nhau nên sửa như vậy nhé)

Vì \(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}\)

\(\Rightarrow\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{2y-4}{6}=\dfrac{3z-9}{12}\)

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

\(\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z-3}{4}=\dfrac{\left(x-1\right)-\left(2y-4\right)+\left(3z-9\right)}{4-6+12}=\dfrac{x-1-2y+4+3z-9}{10}\)

\(=\dfrac{\left(x-2y+3z\right)-\left(1-4+9\right)}{10}=\dfrac{14-6}{10}=\dfrac{4}{5}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x-1}{2}=\dfrac{4}{5}\Rightarrow x=\dfrac{13}{5}\\\dfrac{y-2}{3}=\dfrac{4}{5}\Rightarrow y=\dfrac{22}{5}\\\dfrac{z-3}{4}=\dfrac{4}{5}\Rightarrow z=\dfrac{31}{5}\end{matrix}\right.\)

Vậy \(x=\dfrac{13}{5},\) \(y=\dfrac{22}{5}\) và \(z=\dfrac{31}{5}.\)

c) Giải

Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)

Mà \(x^2+2y^2-z^2=-12\)

\(\Rightarrow\left(2k\right)^2+2\left(3k\right)^2-\left(5k\right)^2=-12\)

\(\Rightarrow4.k^2+18.k^2-25.k^2=-12\)

\(\Rightarrow\left(-3\right)k^2=-12\)

\(\Rightarrow k^2=4\)

\(\Rightarrow k=\pm2\)

\(\circledast k=-2\Rightarrow\left\{{}\begin{matrix}x=-4\\y=-6\\z=-10\end{matrix}\right.\)

\(\circledast k=2\Rightarrow\left\{{}\begin{matrix}x=4\\y=6\\z=10\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=-4;y=-6;z=-10\\x=4;y=6;z=10\end{matrix}\right..\)

câu b bạn ko làm đc hả

a. Đặt \(\dfrac{x}{-3}=\dfrac{y}{5}=k\Leftrightarrow\left\{{}\begin{matrix}x=-3k\\y=5k\end{matrix}\right.\)

mà \(x.y=\dfrac{-5}{27}\)

hay \(-3k.5k=\dfrac{-5}{27}\)

\(\Rightarrow-15.k^2=\dfrac{-5}{27}\)

\(\Rightarrow k^2=\dfrac{1}{81}=\left(\pm\dfrac{1}{9}\right)^2\)

Với \(k=\dfrac{1}{9}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-1}{3}\\y=\dfrac{5}{9}\end{matrix}\right.\)

Với \(k=\dfrac{-1}{9}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=\dfrac{-5}{9}\end{matrix}\right.\)

Vậy.......

b. Từ \(\begin{matrix}\dfrac{x}{3}=\dfrac{y}{4}\\\dfrac{y}{3}=\dfrac{z}{5}\end{matrix}\) \(\Rightarrow\begin{matrix}\dfrac{x}{9}=\dfrac{y}{12}\\\dfrac{y}{12}=\dfrac{z}{20}\end{matrix}\) \(\Rightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)

Áp dụng t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}=\dfrac{x-y+z}{9-12+20}=\dfrac{32}{17}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=\dfrac{32}{17}\Rightarrow x=\dfrac{32.9}{17}=\dfrac{288}{17}\\\dfrac{y}{12}=\dfrac{32}{17}\Rightarrow y=\dfrac{32.12}{17}=\dfrac{384}{17}\\\dfrac{z}{20}=\dfrac{32}{17}\Rightarrow z=\dfrac{32.20}{17}=\dfrac{640}{17}\end{matrix}\right.\)

Vậy.........