Ta có : \(\frac{x}{5}=\frac{y}{5}\Rightarrow\frac{x}{30}=\frac{y}{30}\)   (1)

           \(\frac{y}{6}=\frac{z}{7}\Rightarrow\frac{y}{30}=\frac{z}{35}\) (2)

Từ (1) và (2) \(\Rightarrow\frac{x}{30}=\frac{y}{30}=\frac{z}{35}=k\)

\(\frac{x}{30}=\frac{y}{30}=\frac{z}{35}\Rightarrow x=30k\);  \(y=30k\)và \(z=35k\)

\(x-y+z=23\Rightarrow30k-30k+35k=23\Rightarrow35k=23\Rightarrow k=\frac{23}{35}\)

\(\Leftrightarrow x=30\cdot\frac{23}{35}=\frac{138}{7}\)

       \(y=30\cdot\frac{23}{35}=\frac{138}{7}\)

      \(z=35\cdot\frac{23}{35}=23\)

Vậy .....

\(\frac{x}{5}=\frac{y}{5},\frac{y}{6}=\frac{z}{7}\)

Ta có:\(\frac{x}{30}=\frac{y}{30},\frac{y}{30}=\frac{z}{35}\)

\(\Rightarrow\)\(\frac{x}{30}=\frac{y}{30}=\frac{z}{35}\)

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

\(\frac{x-y+z}{30-30+35}=\frac{23}{35}\)

\(\frac{x}{30}=\frac{23}{35}\Rightarrow x=\frac{138}{7}\)

y =\(\frac{138}{7}\)

z=23

Theo bài ra ta cs 

\(\frac{x}{5}=\frac{y}{5}\Rightarrow\frac{x}{30}=\frac{y}{30}\left(1\right)\)

\(\frac{y}{6}=\frac{z}{7}\Rightarrow\frac{y}{30}=\frac{z}{35}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{x}{30}=\frac{y}{30}=\frac{z}{35}\)

ADTC dãy tỉ số bằng nhau ta cs 

\(\frac{x}{30}=\frac{y}{30}=\frac{z}{35}=\frac{x-y+z}{30-30+35}=\frac{23}{35}\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{30}=\frac{23}{35}\\\frac{y}{30}=\frac{23}{35}\\\frac{z}{35}=\frac{33}{35}\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{138}{7}\\y=\frac{138}{7}\\z=690\end{cases}}}\)

Các bạn giúp mình giải với nhé! Đúng thì mình k đúng nhé. Cảm ơn các bạn nhiều lắm. Yêu cả nhà.

\(1.\left(x-5\right)^{23}.\left(y+2\right)^7=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0\\\left(y+2\right)^7=0\end{cases}\Rightarrow\hept{\begin{cases}\left(x-5\right)^{23}=0^{23}\\\left(y+2\right)^7=0^7\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x-5=0\\y+2=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+5\\y=0-2\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}x=5\\y=-2\end{cases}}\)

Vậy \(\left(x;y\right)=\left(5;-2\right)\)

2. \(A=\left(x-2\right)^2+|y+3|+7\)

Ta có :

\(\hept{\begin{cases}\left(x-2\right)^2\ge0\forall x\\|y+3|\ge0\forall y\end{cases}}\)

\(\Rightarrow\left(x-2\right)^2+|y+3|\ge0\forall x;y\)

\(\Rightarrow\left(x-2\right)^2+|y+3|+7\ge7\forall x;y\)

\(\Rightarrow A\ge7\forall x;y\)

Dấu bằng xảy ra

\(\Leftrightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\|y+3|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x-2=0\\y+3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=-3\end{cases}}}\)

Vậy GTNN của A là 7 khi \(\left(x;y\right)=\left(2;-3\right)\)

1a) \(\frac{x-3}{x+7}=\frac{-5}{-6}\)

=> \(\frac{x-3}{x+7}=\frac{5}{6}\)

=> (x - 3).6 = 5.(x + 7)

=> 6x - 18 = 5x + 35

=> 6x - 5x = 35 + 18

=> x = 53

b) \(\frac{x-7}{x+3}=\frac{4}{3}\)

=> (x - 7). 3 = (x + 3). 4

=> 3x - 21 = 4x + 12

=> 3x - 4x = 12 + 21

=> -x = 33

=> x = -33

c) \(\frac{x-10}{6}=-\frac{5}{18}\)

=> (x - 10) . 18 = -5 . 6

=> 18x - 180 = -30

=> 18x = -30 + 180

=> 18x = 150

=> x = 150 : 18 = 25/3

d) \(\frac{x-2}{4}=\frac{25}{x-2}\)

=> (x - 2)(x - 2) = 25 . 4

=> (x - 2)² = 100

=> (x - 2)² = 10²

=> \(\orbr{\begin{cases}x-2=10\\x-2=-10\end{cases}}\)

=> \(\orbr{\begin{cases}x=12\\x=-8\end{cases}}\)

e) \(\frac{7}{x}=\frac{x}{28}\)

=> 7 . 28 = x . x

=> 196 = x²

=> x² = 14²

=> \(\orbr{\begin{cases}x=14\\x=-14\end{cases}}\)

f) \(\frac{40+x}{77-x}=\frac{6}{7}\)

=> (40 + x) . 7 = (77 - x).6

=> 280 + 7x = 462 - 6x

=> 280 - 462 = -6x + 7x

=> -182 = x

=> x = -182

36367 l'290209 92021

a: \(\dfrac{4}{5}-\dfrac{5}{6}< =\dfrac{x}{30}< =\dfrac{1}{3}-\dfrac{3}{10}\)

=>\(\dfrac{24-25}{30}< =\dfrac{x}{30}< =\dfrac{10-9}{30}\)

=>\(\dfrac{-1}{30}< =\dfrac{x}{30}< =\dfrac{1}{30}\)

=>-1<=x<=1

mà x nguyên

nên \(x\in\left\{-1;0;1\right\}\)

b: \(\dfrac{a}{7}+\dfrac{1}{14}=\dfrac{-1}{b}\)

=>\(\dfrac{2a+1}{14}=\dfrac{-1}{b}\)

=>\(\left(2a+1\right)\cdot b=-14\)

mà 2a+1 lẻ (do a là số nguyên)

nên \(\left(2a+1\right)\cdot b=1\cdot\left(-14\right)=\left(-1\right)\cdot14=7\cdot\left(-2\right)=\left(-7\right)\cdot2\)

=>\(\left(2a+1;b\right)\in\left\{\left(1;-14\right);\left(-1;14\right);\left(7;-2\right);\left(-7;2\right)\right\}\)

=>\(\left(a;b\right)\in\left\{\left(0;-14\right);\left(-1;14\right);\left(3;-2\right);\left(-4;2\right)\right\}\)

.

=> 6x-138-5x-60 = 245

=> x - 198 = 245

=> x = 245 + 198 = 443

k mk nha

a) 

–   5   x – 7   = 20            x − 7        =   20 : − 5            x − 7        = − 4            x                =   − 4 + 7            x                = 3

b)

− 6 x + 2 = 0         x + 2 = 0          x + 2     = 0          x             = − 2