a) 2/x+7=x+7/32

<=> (x+7)^2=64

=> x+7=8 hoặc x+7=-8

=> x=-1 hoặc x=-15

b) - (x+5)^2= (x-2).(x+8)

<=> -(x+5)^2=x^2+8x-2x-16

<=> - (x+5)^2 =(x-4)^2

+> Không có giá trị x thỏa mãn

a. ĐK: x\(\ne\)-7

2.32=(x+7)²

<=> 64=x²+ 14x+ 49

<=>x²+ 14x- 15=0

<=>x²+ 15x- x- 15=0

<=>(x-1)(x+15)=0

\(\Leftrightarrow\hept{\begin{cases}x=1\\x=-15\end{cases}}\)

b, ĐK: x\(\ne\)-5;-8

(x-2)(x+8)=(x-5)(x+5)

<=>x²+ 6x- 16=x²- 25

<=>6x+9=0

\(\Leftrightarrow x=-\frac{3}{2}\)

\(\frac{2}{x+7}=\frac{x+7}{32}\)

\(\Leftrightarrow\left(x+7\right)\left(x+7\right)=64\)

\(\Leftrightarrow\left(x+7\right)^2=8^2\)

\(\Leftrightarrow\orbr{\begin{cases}x+7=8\\x+7=-8\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-15\end{cases}}\)

vậy x=1 hoặc x=-15

b)\(\frac{x-2}{x+5}=\frac{x-5}{x+8}\)ĐKXĐ : \(x\ne-5\) và \(x\ne8\)

\(\Leftrightarrow\left(x+5\right)\left(x-5\right)=\left(x-2\right)\left(x+8\right)\)

\(\Leftrightarrow x^2-25=x^2+6x-16\)

\(\Leftrightarrow6x+9=0\)

\(\Leftrightarrow6x=-9\)

\(\Leftrightarrow x=-\frac{9}{6}=-\frac{3}{2}\)

\(bn\)\(xem\)\(lai\)\(giup\)\(mk\)\(cho\)\(\frac{x+522}{7}\)\(neu\)\(thay\)\(bang\)\(\frac{x+552}{7}\)\(thi\)\(dug\)\(hon\)

thế thì bạn giải thử xem cô t ra đề thế mà ừ thì cứ cho là x + 552 cx đc

Bài 1: \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12};\frac{y}{3}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{20}\)

=>\(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2z}{18}=\frac{3y}{36}\)

Áp dụng tính chất của dãy tỉ số bằng nhau: \(\frac{x}{9}=\frac{y}{12}=\frac{z}{20}=\frac{2z}{18}=\frac{3y}{36}=\frac{2x-3y+z}{18-36+20}=\frac{6}{2}=3\)

=>x=27;z=36;z=60

Bài 2: \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=5k\end{cases}}\Rightarrow xy=2k.5k=10k^2=40\Rightarrow k^2=4\Rightarrow\hept{\begin{cases}k=-2\\k=2\end{cases}}\)

+)k=-2 => x=-4;y=-5

+)k=2 => x=4;y=5

Vậy x=-4;y=-5 hoặc x=4;y=5

\(\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}\)

\(\Rightarrow\frac{x+1}{9}+\frac{x+4}{6}+\frac{x+5}{5}+3=\frac{x+2}{8}+\frac{x+3}{7}+\frac{x+6}{4}+3\)

\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+4}{6}+1\right)+\left(\frac{x+5}{5}+1\right)=\left(\frac{x+2}{8}+1\right)\)\(+\left(\frac{x+3}{7}+1\right)+\left(\frac{x+6}{4}\right)\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)

\(\Rightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}\right)=\left(x+10\right)\left(\frac{1}{8}+\frac{1}{7}+\frac{1}{4}\right)\)

\(\Rightarrow\left(x+10\right)\frac{43}{90}=\left(x+10\right)\frac{29}{56}\)

\(\Rightarrow x+10=0\)

\(\Rightarrow x=-10\)

cộng 3 vào cả hai vế nên phương trình vẫn bằng nhau

Ta có \(\frac{x+1}{9}+1+\frac{x+4}{6}+1+\frac{x+5}{5}+1=\frac{x+2}{8}+1+\frac{x+3}{7}+1+\frac{x+6}{4}+1\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}=\frac{x+10}{8}+\frac{x+10}{7}+\frac{x+10}{4}\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{6}+\frac{x+10}{5}-\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{4}=0\)

\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

mà \(\frac{1}{9}+\frac{1}{6}+\frac{1}{5}-\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\ne0\)

\(\Rightarrow x+10=0\)

\(\Leftrightarrow x=-10\)

tự làm để nc trí óc dễ