Đây nè bạn

a)x-7  = 0 

x=0+7=7

b, ( x - 3 ) . ( x^2 + 3 ) = 0

-> x -3=0  hoặc x^2+3 =0 

+ Nếu x -3 =0 

-> x=3 

+ Nếu x^2+3 =0 

 -> x^2 =-3 ( loại) 

Vậy x=3 

Bài2

6x + 3 chia hết cho x 

 Ta có x chia hết cho x

-> 6x chia hết cho x 

Mà 6x+3 chia hết cho x 

-> (6x+3)-6x chia hết cho x 

-> 3 chia hết cho x

......

Bạn tự làm 

Câu b tương tự

1. 

x - 7 = 0 => x = 7

( x - 3 ) ( x² + 3 ) = 0

=> \(\orbr{\begin{cases}x-3=0\\x^2+3=0\end{cases}}\)=> \(\orbr{\begin{cases}x=3\\x^2=-3\end{cases}}\)

Bình phương một số \(\ge\)0 => x² \(\ne\)-3

=> x = 3

2. a) 6x + 3 chia hết cho x

=> 3 chia hết cho x

=> x thuộc Ư(3) = { -3 ; -1 ; 1 ; 3 }

b) 4x + 4 chia hết cho 2x - 1

=> 2(2x - 1) + 6 chia hết cho 2x - 1

=> 4x - 2 + 6 chia hết cho 2x - 1

=> 6 chia hết cho 2x - 1

=> 2x - 1 thuộc Ư(6) = { -6 ; -3 ; -2 ; -1 ; 1 ; 2 ; 3 ; 6 }

Vì x thuộc Z => x thuộc { -1 ; 0 ; 1 ; 2 }

a: (x+2)(x-3)>0

nên x+2;x-3 cùng dấu

=>x>3 hoặc x<-2

b: (x-1)(x+4)<=0

nên x-1 và x+4 khác dấu

=>-4<=x<=1

a: f(-2)=4+3=7

f(-1)=2+3=5

f(0)=3

f(1/2)=-1+3=2

f(-1/2)=1+3=4

b: g(-1)=1-1=0

f(0)=0-1=-1

a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)

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

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)

b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)

hay \(x=-\dfrac{1}{4}\)

c) Ta có: \(8x^3-50x=0\)

\(\Leftrightarrow2x\left(4x^2-25\right)=0\)

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

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

e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)

f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)

a. Ta có: \(x^2-10x+26+y^2+2y=0\Leftrightarrow\left(x^2-10x+25\right)+\left(y^2+2y+1\right)=0\\ \)

\(\Leftrightarrow\left(x+5\right)^2+\left(y+1\right)^2=0\Rightarrow\hept{\begin{cases}x+5=0\\y+1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\y=-1\end{cases}}}\)

b. \(\left(2x+5\right)^2-\left(x-7\right)^2=0\Leftrightarrow\left(2x+5+x-7\right).\left(2x+5-x+7\right)=0\)

\(\Leftrightarrow\left(3x-2\right).\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}3x-2=0\\x+12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-12\end{cases}}}\)

c. \(25.\left(x-3\right)^2=49.\left(1-2x\right)^2\Leftrightarrow\left(5x-15\right)^2=\left(7-14x\right)^2\Leftrightarrow\left(5x-15\right)^2-\left(7-14x\right)^2=0\)

\(\Leftrightarrow\left(5x-15-7+14x\right).\left(5x-15+7-14x\right)=0\Leftrightarrow\left(19x-22\right).\left(-9x-8\right)=0\)

\(\Leftrightarrow\left(19x-22\right).\left(9x+8\right)=0\Leftrightarrow\orbr{\begin{cases}19x-22=0\\9x+8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{22}{19}\\x=-\frac{8}{9}\end{cases}}}\)

d. \(\left(x+2\right)^2=\left(3x-5\right)^2\Leftrightarrow\left(x+2\right)^2-\left(3x-5\right)^2=0\Leftrightarrow\left(x+2+3x-5\right).\left(x+3-3x+5\right)=0\)

\(\Leftrightarrow\left(4x-3\right).\left(8-2x\right)=0\Leftrightarrow\orbr{\begin{cases}4x-3=0\\8-2x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=4\end{cases}}}\)

e. \(x^2-2x+1=16\Leftrightarrow\left(x-1\right)^2-16=0\Leftrightarrow\left(x-1-4\right).\left(x-1+4\right)=0\)

\(\Leftrightarrow\left(x-5\right).\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x-5=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)

Cảm ơn bn rất nhìu nha!!!^-^!!!

Bn ơi cho mik hỏi cái này đc ko ở câu a hai số cộng vs nhau bằng 0 thì một trong hai số đó là băng 0 à bn . Hi hi mik lại nghĩ là hai số nhân vs nhau bằng 0 thì 1 trong hai số đó ms bằng 0 chứ!!! Dù sao thì cx cảm ơn bn nhìu nà!!! Chúc bn học tốt nha!!!

a. \(x^2-25-3.\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5\right)-3.\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+5-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

b. \(\left(3x+1\right)^2=\left(2x-5\right)\\ \Leftrightarrow9x^2+6x+1=2x-5\\ \Leftrightarrow9x^2+6x-2x=-5-1\\ \Leftrightarrow9x^2+4x=-6\\ \Leftrightarrow x\left(9x+4\right)=-6\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\9x+4=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-\dfrac{10}{9}\end{matrix}\right.\)

c. \(2x^2-7x+6=0\\ \Leftrightarrow2x^2-7x=-6\\ \Leftrightarrow x\left(2x-7\right)=-6\\ \Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\dfrac{1}{2}\end{matrix}\right.\)

a, \(\left(x-5\right)\left(x+5\right)-3\left(x-5\right)=0\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\Leftrightarrow x=-2;x=5\)

b, bạn ktra lại đề, thường thường ngta hay cho 2 vế cùng bình phương 

c, \(2x^2-7x+6=0\Leftrightarrow\left(2x-3\right)\left(x-2\right)=0\Leftrightarrow x=\dfrac{3}{2};x=2\)

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

\(\left[{}\begin{matrix}3x+1=2x-5\\3x+1=5-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\5x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=\dfrac{4}{5}\end{matrix}\right.\)

\(a,4x\left(x+1\right)=8\left(x+1\right)\)

\(\Leftrightarrow4x^2+4x-8x-8=0\)

\(\Leftrightarrow4x^2-4x-8=0\)

\(\Leftrightarrow4\left(x^2-x-2\right)=0\)

\(\text{⇔}4\left(x^2-2x+x-2\right)=0\)

\(\text{⇔}4\left(x-2\right)\left(x+1\right)=0\)

\(\text{⇔}\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

\(c,2x\left(x-2\right)-\left(2-x\right)^2=0\)

\(\text{⇔}2x\left(x-2\right)-\left(x-2\right)^2=0\)

\(\text{⇔}\left(x-2\right)\left(2x-x+2\right)=0\)

\(\text{⇔}\left(x-2\right)\left(x+2\right)=0\)

\(\text{⇔}\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

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

\(\text{⇔}\left(x-3\right)^3-\left(x-3\right)=0\)

\(\text{⇔}\left(x-3\right)\left(x^2-6x+9-1\right)=0\)

\(\text{⇔}\left(x-3\right)\left(x^2-6x+8\right)=0\)

\(\text{⇔}\left(x-3\right)\left(x-2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=2\\x=4\end{matrix}\right.\)

\(g,5x\left(x-2000\right)-x+2000=0\)

\(\text{⇔}\left(x-2000\right)\left(5x-1\right)=0\)

\(\text{⇔}\left[{}\begin{matrix}x=2000\\x=\frac{1}{5}\end{matrix}\right.\)

\(n,\left(x+1\right)^2-1+x=0\)

\(\text{⇔}x^2+2x+1-1+x=0\)

\(\text{⇔}x^2+3x=0\)

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

\(k,\left(1-x\right)^2-1+x=0\)

\(\text{⇔}\left(1-x\right)^2-\left(1-x\right)=0\)

\(\text{⇔}\left(1-x\right)\left(1-x-1\right)=0\)

\(\text{⇔}\left(1-x\right).\left(-x\right)=0\)

\(\text{⇔}\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)

\(m,x+6x^2=0\)

\(\text{⇔}x\left(1+6x\right)=0\)

\(\text{⇔}\left[{}\begin{matrix}x=0\\x=-\frac{1}{6}\end{matrix}\right.\)

\(h,x^2-4x=0\)

\(\text{⇔}x\left(x-4\right)=0\)

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