17 + (x - 7) + 2(2x + 1) = 17

<=> x - 7 + 2(2x + 1) = 17 - 17

<=> x - 7 + 2(2x + 1) = 0

<=> x + 2(2x + 1) = 0 + 7

<=> x + 2(2x + 1) = 7

<=> x + 4x + 1 = 7

<=> 5x + 2 = 7

<=> 5x = 7 - 2

<=> 5x = 5

=> x = 1

\(17+\left(x-7\right)+2\left(2x+1\right)=17\)

\(\Rightarrow x-7+4x+2=0\)

\(\Rightarrow5\left(x-1\right)=0\)

\(\Rightarrow x-1=0\)

\(\Rightarrow x=1\)

\(17+\left(x-7\right)+2\left(2x+1\right)=17\)

\(\Leftrightarrow17+x-7+4x+2=17\)

\(\Leftrightarrow\left(17-7\right)+\left(x+2x\right)+2=17\)

\(\Leftrightarrow10+3x+2=17\)

\(\Leftrightarrow\left(10+2\right)+3x=17\)

\(\Leftrightarrow12+3x=17\)

\(\Leftrightarrow3x=17-12\)

\(\Leftrightarrow3x=5\)

\(\Leftrightarrow x=\frac{5}{3}\)

\(\frac{8}{5}=\frac{-12}{x}\left(x\ne0\right)\)\(\Leftrightarrow8x=-60\)\(\Leftrightarrow x=\frac{-60}{8}=\frac{-15}{2}\)(tmđk)

\(\frac{x-1}{-4}=\frac{-4}{x-1}\left(x\ne1\right)\)\(\Leftrightarrow\left(x-1\right)^2=16\)\(\Leftrightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\left(tm\right)\\x=-3\left(tm\right)\end{cases}}}\)

\(\frac{8}{5}=\frac{-12}{x}\)

\(\Rightarrow8x=-60\)

        \(x=-60:8\)

        \(x=-7,5\)

Vậy x=-7,5

\(\frac{x-1}{-4}=\frac{-4}{x-1}\)

\(\Rightarrow\left(x-1\right)^2=16\)

     \(\left(x-1\right)^2=4^2\)

 \(\Rightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=4+1=5\\x=-4+1=-3\end{cases}}\)

vậy x=5 hoặc x=-3

Bài làm

a) 8/5 = -12/x

=> x = -12.5/8 = -7,5

Vậy x = -7,5

b) x-1/-4 = -4/x-1

=> ( x - 1 )( x - 1 ) = -4 . (-4)

=> x² - 2x + 1 = 16

=> x² - 2x + 1 - 16 = 0

=> x² - 2x - 15 = 0

=> x² - 5x + 3x - 15 = 0

=> x( x - 5 ) + 3( x - 5 ) = 0

=> ( x - 5 )( x + 3 ) = 0

=> x - 5 = 0 hoặc x + 3 = 0

=> x = 5 hoặc x = -3

Vậy x = 5 hoặc x = -3

c) -2/x = y/4

=> xy = -8

d) 4/y+2 = 7/3y+1 

=> 4( 3y + 1 ) = 7( y + 2 )

=> 12y + 4 = 7y + 14

=> 5y = 10

=> y = 2

Vậy y = 2

e) 2x-1/3 = 3x+1/4

=> 4( 2x-1) = 3( 3x + 1 )

=> 8x - 4 = 9x + 3

=> -x = 7

=> x = -7

Vậy x = -7

\(-2x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x-4=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

\(-2x\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

a) \(10⋮2x+1\)

\(\Rightarrow2x+1\inƯ\left(10\right)=\left\{\pm1;\pm2;\pm5;\pm10\right\}\)

Xét bảng sau:

a) \(A=\)\(x^4\)\(+4x^3\)\(+2x^2\)\(+x\)\(-7\)

  \(B=\)\(2x^4\)\(-4x^3\)\(-2x^2\)\(-5x\)\(+3\)

b) f(x)= A(x)+B(x)= \(3x^4-4x\)\(-4\)

    g(x)=A(x)-B(x) =  \(-x^4+8x^3+4x^2+6x\)\(-10\)

c) g(x)= \(0^4+8.0^3+4.0^2\)\(+6.0\)\(-10\)

         = -10

   g(-2)=\(-2^4+8.-2^3+4.-2^2+6.-2\)\(-10\)

         =\(-54\)

2x=3y

=>\(\dfrac{x}{3}=\dfrac{y}{2}\)

=>\(\dfrac{x}{9}=\dfrac{y}{6}\)

4y=3z

=>\(\dfrac{y}{3}=\dfrac{z}{4}\)

=>\(\dfrac{y}{6}=\dfrac{z}{8}\)

=>\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}\)

mà x-y+2z=57

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

\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{x-y+2z}{9-6+2\cdot8}=\dfrac{57}{19}=3\)

=>x=27; y=18; z=24

Theo bài ra, ta có: \(2x-1⋮x-1\)

\(\Rightarrow2\left(x-1\right)+1⋮x-1\)

\(\Rightarrow1⋮x-1\)

Vì \(x\in Z\Rightarrow x-1\inƯ\left(1\right)=\left\{\mp1\right\}\)

Ta có các trường hợp sau:

\(\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)

Vậy \(x\in\left\{2;0\right\}\)

2x-1 chia hết cho x-1

=>2x-2+1 chia hết cho x-1

=>2(x-1)+1 chia hết cho x-1

mà 2(x-1) chia hết cho x-1

=>1 chia hết cho x-1 => x-1 thuộc Ư(1) thuộc {1;-1} => x thuộc {2;0}

Vậy x thuộc {2;0}

Chúc bạn học tốt ^^!!!