a,                                          b,                                             c,

7x-8/6x+5=0                           7x-8/6x+5>0                             7x-8/6x+5<0     

7x-8=0                                   7x-8>0                                      7x-8<0

x=8/7                                     x>8/7                                        x<8/7

ĐÚNG RỒI ĐÓ, nhớ k cho mk nha

7x = 4y nên \(\frac{y}{\frac{1}{4}}=\frac{x}{\frac{1}{7}}=\frac{y-x}{\frac{1}{4}-\frac{1}{7}}=\frac{24}{\frac{3}{28}}=224\)=> x = 224 : 7 = 32 ; y = 224 : 4 = 56

7x=4y => y/7 = x/4

ap dung day ty so = nhau ta co;

(y-x) /(7-4) = 24/3 =8

x= 4.8 = 32

y = 7.8 = 56

Theo mình thì bạn Phan Thanh Tịnh làm đúng đó!

Ta có: |x+1| ;|x+3| ;|x+5|>=0

=> |x+1|+|x+3|+|x+5|>=0

=> 7x>=0

=> x+1+x+3+x+5=7x

      3x+8=7x

        4x=8

x=2

|x + 1| + |x + 3| + |x + 5| = 7x

có |x + 1| > 0; |x + 3| > 0; |x + 5| > 0 

=> 7x > 0

=> x > 0

=> x + 1 + x + 3 + x + 5 = 7x

=> 3x + 9 = 7x

=> 3x - 7x = - 9

=> -4x = -9

=> x = 9/4

\(\left|x+1\right|+\left|x+3\right|+\left|x+5\right|=7x\)

Vì \(\left|x\right|\ge0\)nên \(\left|x+1\right|\ge0;\left|x+3\right|\ge0;\left|x+5\right|\ge0\)

Mà \(7x\ge0\)

\(\Rightarrow x+1+x+3+x+5=7x\)

\(\Rightarrow3x+9=7x\)

\(\Rightarrow7x-3x=9\)

\(\Rightarrow4x=9\)

\(\Rightarrow x=\frac{9}{4}\)

a)

 \(\left(2x-15\right)^5=\left(2x-15\right)^3\\ \Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\\ \Leftrightarrow\left(2x-15\right)^3.\left[\left(2x-15\right)^2-1\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15-1\right).\left(2d-15+1\right)=0\end{matrix}\right.\\\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right. \)

b) \(\left(7x-11\right)^3=\left(-3\right)^2.15+208\\ \Leftrightarrow\left(7x-11\right)^3=343=7^3\\ \Leftrightarrow7x-11=7\\ \Leftrightarrow x=\dfrac{18}{7}\)

\(\text{a. }-15+x=-4\)

\(\Rightarrow x=-4+15\)

\(\Rightarrow x=11\)

\(\text{Vậy }x=11\)

\(\text{b.}35-x=-12-3\)

\(\Rightarrow35-x=-15\)

\(\Rightarrow x=35+15\)

\(\Rightarrow x=50\)

\(\text{Vậy }x=50\)

\(\text{c.}11x-7x+x=325\)

\(\Rightarrow5x=325\)

\(\Rightarrow x=325\div5\)

\(\Rightarrow x=65\)

\(\text{Vậy }x=65\)

\(\text{d.}15-\left|x-3\right|=7\)

\(\Rightarrow\left|x-3\right|=15-7\)

\(\Rightarrow\left|x-3\right|=8\)

\(\Rightarrow\orbr{\begin{cases}x-3=-8\\x-3=8\end{cases}}\)

\(\text{Trường hợp : }x-3=-8\)

\(\Rightarrow x=-8+3\)

\(\Rightarrow x=-5\)

\(\text{Trường hợp : }x-3=8\)

\(\Rightarrow x=8+3\)

\(\Rightarrow x=11\)

\(\text{Vậy }x\in\left\{-5;11\right\}\)

a,-15+x=-4                   b,35-x=-12-3

x=-4+15                          35-x=-12+(-3)

x=11                                  35-x=-15

vậy x=11                                 x=35+15

                                                x=50

                                                   vẫy=50

a.

-2 (x+6)+6 (x-10)=8

-2x -12 +6x -60 =8

4x -72 =8

4x = -64

  x= -16

b.

7x (2+x) -7x (x+3) =14

7x [ (2+x)-(x+3) ] =14

7x (2+x-x-3) =14

-7x =14

x= -2

\(1,\Rightarrow3^{x-3}=\left(3^2\right)^8:\left(3^3\right)^5=3^{16}:3^{15}=3^1\\ \Rightarrow x-3=1\\ \Rightarrow x=4\\ 2,\Rightarrow7^x\left(1+7^2\right)=350\\ \Rightarrow7^x=\dfrac{350}{50}=7=7^1\\ \Rightarrow x=1\)

\(3,\Rightarrow2^{2+2x+2}-2^{2x}=240\\ \Rightarrow2^{2x}\left(2^4-1\right)=240\\ \Rightarrow2^{2x}=\dfrac{240}{15}=16=2^4\\ \Rightarrow2x=4\Rightarrow x=2\)

Bài 1:

1)    \(\left|x-15\right|+x-15=0\)\(\Leftrightarrow\)\(\left|x-15\right|=15-x\)

 + Với \(x\ge15\forall x\)\(\Leftrightarrow\)\(x-15\ge0\forall x\)\(\Rightarrow\)\(\left|x-15\right|=x-15\)

  \(\Rightarrow x-15=15-x\)

 \(\Leftrightarrow2x=30\)

 \(\Leftrightarrow x=15\)( thỏa mãn điều kiện )

 + Với \(x< 15\forall x\)\(\Leftrightarrow\)\(x-15< 0\forall x\)\(\Rightarrow\)\(\left|x-15\right|=-\left(x-15\right)=15-x\)

  \(\Rightarrow15-x=15-x\)

 \(\Leftrightarrow0x=0\)( Vô số các giá trị. Điều kiện: \(x< 15\))

Vậy \(x\le15\)

2)   \(7x.\left(2+x\right)-7x.\left(x+3\right)=14\)

\(\Leftrightarrow7x.\left(2+x-x-3\right)=14\)

\(\Leftrightarrow-7x=14\)

\(\Leftrightarrow x=-2\)( thỏa mãn )

Vậy \(x=-2\)

Bài 2:

1) Ta có: \(A=-3x^3-2x^2+x-14\)

        \(\Leftrightarrow A=-\left(3x^3+6x^2\right)+\left(4x^2+8x\right)-\left(7x+14\right)\)

        \(\Leftrightarrow A=-3x^2.\left(x+2\right)+4x.\left(x+2\right)-7.\left(x+2\right)\)

        \(\Leftrightarrow A=\left(x+2\right).\left(-3x^2+4x-7\right)\)

 + Thay \(x=-3\)vào biểu thức A, ta có:

           \(A=\left(-3+2\right).\left(-3.9-12-7\right)\)

    \(\Leftrightarrow A=\left(-1\right).\left(-46\right)\)

    \(\Leftrightarrow A=46\)

 Vậy \(A=46\)

2) Ta có: \(B=2xy-3x+2y\)

 + Thay \(x=-2,x=-5\)vào biểu thức B, ta có:

            \(B=2.\left(-2\right).\left(-5\right)-3.\left(-2\right)+2.\left(-5\right)\)

     \(\Leftrightarrow B=20+6-10\)

     \(\Leftrightarrow B=16\)

 Vậy \(B=16\) 

Ta có: \(\dfrac{3-x}{20}=\dfrac{-5}{x-2}\)

\(\Leftrightarrow\dfrac{x-3}{-20}=\dfrac{-5}{x-2}\)

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=100\)

\(\Leftrightarrow x^2-5x+6-100=0\)

\(\Leftrightarrow x^2-5x-94=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5-\sqrt{401}}{2}\\x=\dfrac{5+\sqrt{401}}{2}\end{matrix}\right.\)