mà đây là môn toán mà

1.a\(\ne\)0

3.a\(\ne\)3

7.a\(\ne\)1

l x + 1 l - 5 = 0

l x + 1 l      = 0 + 5 = 5

  x + 1 = 5

  x       = 5 - 1 = 4

x + 1 = -5

x       = -5 - 1 = - 6

a: \(x^2-\dfrac{3}{2}=0\)

nên \(x^2=\dfrac{3}{2}\)

hay \(x\in\left\{\dfrac{\sqrt{6}}{2};-\dfrac{\sqrt{6}}{2}\right\}\)

b: \(\dfrac{1}{2}x^2+\dfrac{7}{2}x=0\)

\(\Leftrightarrow x^2+7x=0\)

=>x(x+7)=0

=>x=0 hoặc x=-7

c: \(2x\left(x-\dfrac{1}{7}\right)=0\)

=>x(x-1/7)=0

=>x=0 hoặc x=1/7

d: (3x-2)(2x-2/3)=0

=>3x-2=0 hoặc 2x-2/3=0

=>3x=2 hoặc 2x=2/3

=>x=2/3 hoặc x=1/3

a)

\(\left(x+2\right)^2-9=0\)

\(\Rightarrow\left(x+2\right)^2=9=3^2\)

\(\Rightarrow x+2=\pm3\)

\(\Rightarrow x=-5;1\)

b)

\(25x^2-10x+1=0\)

\(\left(5x\right)^2-2\cdot5x+1^2=0\)

\(\Rightarrow\left(5x+1\right)^2=0\)

\(\Rightarrow5x+1=0\)

\(\Rightarrow5x=-1;x=\dfrac{-1}{5}\)

c)

\(x^2+14x+49=0\)

\(\Rightarrow x^2+2\cdot7x+7^2=0\)

\(\Rightarrow\left(x+7\right)^2=0;x+7=0\)

\(\Rightarrow x=-7\)

d)

\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(4x^2-4x+1+x^2+6x+9-5x^2+5\cdot49=0\)

\(\Rightarrow5x^2-5x^2-4x+6x+10+245=0\)

\(\Rightarrow2x+255=0\)

\(\Rightarrow2x=-255\)

\(\Rightarrow x=\dfrac{-255}{2}\)

1) tìm x : 

5x. (x - 3 ) + 7.(x - 3 ) = 0

<=> ( x -3 ) . ( 5x +7 ) = 0

<=> x - 3 = 0 hoặc 5x + 7 = 0 

<=> x = 3 hoặc x = -7/5

Vậy x € { 3 ; -7/5 }

3 ) chứng mình rằng : 

7 ¹⁹⁹⁶ + 7¹⁹⁹⁵ + 7¹⁹⁹⁴ chia hết cho 57 

7¹⁹⁹⁶ + 7¹⁹⁹⁵ + 7¹⁹⁹⁴ 

<=> 7¹⁹⁹⁴  . 7² + 7¹⁹⁹⁴ .7 + 7¹⁹⁹⁴

<=> 7¹⁹⁹⁴ . ( 7²  + 7 + 1 ) 

<=> 7¹⁹⁹⁴ .  57 chia  hết cho 57 ( vì 57 chia hết cho 57 )  ( đ..p.c.m ) 

Bài 1 : \(5x\left(x-3\right)+7\left(x-3\right)=0.\)

\(\Rightarrow5x^2-15x+7x-21=0\)

\(\Rightarrow5x^2-8x-21=0\)

\(\Rightarrow5x^2-15x+7x-21=0\)

\(\Rightarrow5x\left(x-3\right)+7\left(x-3\right)=0\)

\(\Rightarrow\left(x-3\right)\left(5x-7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-3=0\\5x-7=0\end{cases}\Rightarrow\hept{\begin{cases}x=3\\x=\frac{7}{5}\end{cases}}}\)

Bài 2 : \(a,A=0\Rightarrow x^2-3x=0\Rightarrow x\left(x-3\right)=0\Rightarrow x\in\left\{0;3\right\}\)

\(b,A>0\Rightarrow x^2-3x>0\Rightarrow x\left(x-3\right)>0\)

TH1 : \(\hept{\begin{cases}x>0\\x-3>0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x>3\end{cases}\Rightarrow}x>3}\)

TH2 : \(\hept{\begin{cases}x< 0\\x-3< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 0\\x< 3\end{cases}\Rightarrow}x< 3}\)

C, tương tự 

Bài 3 : \(7^{1996}+7^{1995}+7^{1994}=7^{1994}\left(7^2+7+1\right)\)

\(=7^{1994}.57\)\(⋮\)\(7\)

\(\Rightarrow7^{1996}+7^{1995}+7^{1994}⋮\)\(7\)

Bài làm:

Bài 1:

\(5x\left(x-3\right)+7\left(x-3\right)=0\)

\(\Leftrightarrow\left(5x+7\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5x+7=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=-7\\x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{7}{5}\\x=3\end{cases}}}\)

Bài 2: 

a) \(A=0\Leftrightarrow x^2-3x=0\Leftrightarrow x\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}}\)

b) \(A>0\Rightarrow x^2-3x>0\Leftrightarrow x\left(x-3\right)>0\)

\(\Rightarrow\orbr{\begin{cases}x>3\\x< 0\end{cases}}\)

c) \(A< 0\Rightarrow x^2-3x< 0\Leftrightarrow x\left(x-3\right)< 0\)

\(\Rightarrow0< x< 3\)

Bài 3:

Ta có: \(7^{1996}+7^{1995}+7^{1994}=7^{1994}\left(7^2+7+1\right)=7^{1994}.57⋮57\)

\(\Rightarrowđpcm\)

Câu 1 : \(-a.\left(c-d\right)-d.\left(a+c\right)=-c.\left(a+d\right)\)

Ta có : \(VT=-a.\left(c-d\right)-d\left(a+c\right)\)

                 \(=-ac+ad-da-dc\)

                 \(=-ac-dc\)

                 \(=-c\left(a+d\right)=VP\)

\(\Rightarrow-a\left(c-d\right)-d\left(a+c\right)=-c\left(a+d\right)\left(đpcm\right)\)

Câu 2 : 

1,  \(x.\left(x+7\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=-7\end{cases}}}\)

2, \(\left(x+12\right)\left(x-3\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+12=0\\x-3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-12\\x=3\end{cases}}}\)

3, \(\left(-x+5\right)\left(3-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}-x+5=0\\3-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=3\end{cases}}}\)

4, \(x\left(2+x\right)\left(7-x\right)=0\)

\(\Rightarrow x=0;2+x=0\)hoặc \(7-x=0\)

\(\Rightarrow x=0;x=-2\)hoặc \(x=7\)

Thanks Bạn!!

3:

a: =>x=0 hoặc x+5=0

=>x=0 hoặc x=-5

b: =>x^2=4

=>x=2 hoặc x=-2

c: =>(x-5)(2x+1+x+6)=0

=>(x-5)(3x+7)=0

=>x=5 hoặc x=-7/3

1.

a. 2x - 6 > 0 

\(\Leftrightarrow\)  2x  > 6

\(\Leftrightarrow\)    x  > 3

S = \(\left\{x\uparrow x>3\right\}\) 

b. -3x + 9 > 0

\(\Leftrightarrow\)  - 3x   > - 9 

\(\Leftrightarrow\)      x < 3

S = \(\left\{x\uparrow x< 3\right\}\) 

c. 3(x - 1) + 5 > (x - 1) + 3

\(\Leftrightarrow\) 3x - 3 + 5 > x - 1 + 3

\(\Leftrightarrow\) 3x - 3 + 5 - x + 1 - 3 > 0

\(\Leftrightarrow\) 2x > 0 

\(\Leftrightarrow\)   x > 0

S = \(\left\{x\uparrow x>0\right\}\) 

d. \(\dfrac{x}{3}-\dfrac{1}{2}>\dfrac{x}{6}\) 

\(\Leftrightarrow\dfrac{2x}{6}-\dfrac{3}{6}>\dfrac{x}{6}\)

\(\Leftrightarrow2x-3>x\)

\(\Leftrightarrow2x-3-x>0\)

\(\Leftrightarrow x-3>0\)

\(\Leftrightarrow x>3\)

\(S=\left\{x\uparrow x>3\right\}\)

2.

a. 

Ta có: a > b

3a > 3b (nhân cả 2 vế cho 3)

3a + 7 > 3b + 7 (cộng cả 2 vế cho 7)

b. Ta có: a > b

a > b (nhân cả 2 vế cho 1)

a + 3 > b + 3 (cộng cả 2 vế cho 3) (1)

Ta có; 3 > 1

b + 3 > b + 1 (nhân cả 2 vế cho 1b) (2)

Từ (1) và (2) \(\Rightarrow\) a + 3 > b + 1 

c.

5a - 1 + 1 > 5b - 1 + 1 (cộng cả 2 vế cho 1)

5a . \(\dfrac{1}{5}\) > 5b . \(\dfrac{1}{5}\) (nhân cả 2 vế cho \(\dfrac{1}{5}\) )

a > b

3.

a. 2x(x + 5) = 0

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\) 

\(S=\left\{0,-5\right\}\)

b. x² - 4 = 0 

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

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

\(S=\left\{0,4\right\}\)

d. (x - 5)(2x + 1) + (x - 5)(x + 6) = 0

\(\Leftrightarrow\left(x-5\right)\left(2x+1+x+6\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(3x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-7}{3}\end{matrix}\right.\)

\(S=\left\{5,\dfrac{-7}{3}\right\}\)