\(\frac{6}{x+27}=-\frac{7}{x+1}\)

\(\Rightarrow6\left(x+1\right)=-7\left(x+27\right)\)

\(6x+6=-7x+\left(-189\right)\)

\(6x+7x=-189-6\)

\(13x=195\)

     \(x=195:13\)

    \(x=15\)

Vậy \(x=15\)

Ta có:   \(\frac{6}{x+27}=\frac{-7}{x+1}\)

       \(\Leftrightarrow6\cdot\left(x+1\right)=-7\cdot\left(x+27\right)\)

       \(\Leftrightarrow6x+6=-7x-189\)

       \(\Leftrightarrow6x+7x=-189-6\)

       \(\Leftrightarrow13x=-195\)

       \(\Leftrightarrow x=-15\)

 Vậy \(x=-15\)

      \(\approx GOOD\)\(LUCK\approx\)

c: \(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=-\sqrt{7}\\x=-5\\x=5\end{matrix}\right.\)

`(x-7)^{2}-(x-2)^{2}=26`

`<=>(x-7-x+2)(x-7+x-2)=26`

`<=>-5(2x-9)=26`

`<=>2x-9=-26/5`

`<=>2x=19/5`

`<=>x=19/10`

\(\left(x-7\right)^2-\left(x-2\right)^2=26\)

\(< =>x^2-14x+49-\left(x-2\right)^2=26\)

\(< =>x^2-14x+49-\left(x^2-4x+4\right)=26\)

\(< =>x^2-14x+49-x^2+4x-4=26\)

\(< =>-10x+49=26+4\)

\(< =>-10x=30-49\)

\(< =>-10x=-19\)

\(< =>x=\dfrac{19}{10}\)

`(x-7)^{2}-(x-2)^{2}=26`

`<=>(x-7-x+2)(x-7+x-2)=26`

`<=>-5(2x-9)=26`

`<=>2x-9=-26/5`

`<=>2x=71/5`

`<=>x=71/10`

Mk ko viết lại đề bài nhé

<=> -2-2-2-...-2=-100

<=>(-2)(\({\\{x-1} \over 4}\)+1)=-100

<=>\({\\{x-1} \over 4}\)+1=50

<=>\({\\{x-1} \over 4}\)=49

<=>x-1=196

<=>x=197

Nhớ k nha

x = 197

a) \(\dfrac{x+1}{32}=\dfrac{2}{x+1}\)

\(\Leftrightarrow\dfrac{x+1}{32}=\dfrac{2}{x+1}\left(đk:x\ne1\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x+1\right)=64\)

\(\Leftrightarrow\left(x+1\right)^2-64=0\)

\(\Leftrightarrow x^2+2x+1-64=0\)

\(\Leftrightarrow x^2+6x-63=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+16}{2}\\x=\dfrac{-2-16}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-9\end{matrix}\right.\left(đk:x\ne-1\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-9\end{matrix}\right.\)

Vậy \(x_1=-9;x_2=7\)

b) \(\dfrac{x+1}{5}=\dfrac{7}{x-1}\)

\(\Leftrightarrow\dfrac{x+1}{5}=\dfrac{7}{x-1}\left(đk:x\ne1\right)\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=35\)

\(\Leftrightarrow\left(x+1\right)\left(x-1\right)-35=0\)

\(\Leftrightarrow x^2-1-35=0\)

\(\Leftrightarrow x^2-36=0\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\left(đk:x\ne1\right)\)

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

Vậy \(x_1=-6;x_2=6\)

c) \(\left|4,5-2x\right|:1\dfrac{7}{4}=\dfrac{11}{14}\)

\(\Leftrightarrow\left|4,5-2x\right|:\dfrac{11}{4}=\dfrac{11}{4}\)

\(\Leftrightarrow\left|4,5-2x\right|\cdot\dfrac{4}{11}=\dfrac{11}{14}\)

\(\Leftrightarrow\dfrac{4}{11}\cdot\left|4,5-2x\right|=\dfrac{11}{14}\)

\(\Leftrightarrow\left|4,5-2x\right|=\dfrac{121}{56}\)

\(\Leftrightarrow\left[{}\begin{matrix}4,5-2x=\dfrac{121}{56}\\4,5-2x=-\dfrac{121}{56}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{131}{112}\\x=\dfrac{373}{112}\end{matrix}\right.\)

Vậy \(x_1=\dfrac{131}{112};x_2=\dfrac{373}{112}\)

a) \(\dfrac{x+1}{32}=\dfrac{2}{x+1}\)

\(\Rightarrow\left(x+1\right)\left(x+1\right)=32.2\)

\(\Rightarrow\left(x+1\right)^2=64\)

\(\Rightarrow\left(x+1\right)^2=8^2\)

\(\Rightarrow x+1=8\)

\(\Rightarrow x=8-1\)

\(\Rightarrow x=7\left(TM\right)\)

Vậy \(x=7\) là giá trị cần tìm

b) \(\dfrac{x+1}{5}=\dfrac{7}{x-1}\)

\(\Rightarrow\left(x+1\right)\left(x-1\right)=7.5\)

\(\Rightarrow\left[{}\begin{matrix}x+1=7\\x-1=5\end{matrix}\right.\) \(\Rightarrow x=6\left(TM\right)\)

Vậy \(x=6\) là giá trị cần tìm

c) \(\left|4,5-2x\right|:1\dfrac{7}{4}=\dfrac{11}{14}\)

\(\left|\dfrac{45}{10}-2x\right|:\dfrac{11}{4}=\dfrac{11}{4}\)

\(\left|\dfrac{9}{2}-2x\right|=\dfrac{11}{14}.\dfrac{11}{4}\)

\(\left|\dfrac{9}{2}-2x\right|=\dfrac{121}{56}\)

\(\Rightarrow\left[{}\begin{matrix}\dfrac{9}{2}-2x=\dfrac{121}{56}\\\dfrac{9}{2}-2x=\dfrac{-121}{56}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=\dfrac{131}{56}\\2x=\dfrac{373}{56}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{131}{112}\\x=\dfrac{373}{112}\end{matrix}\right.\)

Vậy \(x\in\left\{\dfrac{131}{112};\dfrac{373}{112}\right\}\) là giá trị cần tìm

b) x +1/5 = 7/x - 1

Suy ra: ( x+1).(x - 1) = 7.5

Nếu x +1= 7 Nếu x -1 = 5

X = 6 x = 6

Vậy x cần tìm là 6

Bạn ơi đề bài ko có y

(x - 7)^x + 1 - (x - 7)^x + 11 = 0

\(\Rightarrow\) [(x - 7)^x - (x - 7)^x] + (11 + 1) = 0

\(\Rightarrow\) 0 + 12 = 0

\(\Rightarrow\) 12 = 0

\(\Rightarrow\) x \(\in\) {\(\phi\)}

Vậy không có giá trị x nào để (x - 7)^x + 1 - (x - 7)^x + 11 = 0

Chúc bạn học tốt!

mày ko ko bik dùng máy tính giải tìm x ak con bitch

a) x - 15 = 11 - (-32)

x - 15 = 43

x= 43+15

x=58

Vậy...

b) 13 - (5-x) = 7

5-x= 13-7

5-x= 6

x= 5-6

x= -1

Vậy...