Bài 3:

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

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

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

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

Vậy.......................

2. \(\left(3x+5\right)\left(x-3\right)-6x-10=0\)

\(\Rightarrow\left(3x+5\right)\left(x-3\right)-2\left(3x+5\right)=0\)

\(\Rightarrow\left(3x+5\right)\left(x-3-2\right)=0\)

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

Vậy........................

3. \(\left(x-2\right)\left(2x+3\right)-7x^2+14x=0\)

\(\Rightarrow\left(x-2\right)\left(2x+3\right)-7x\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(2x+3-7x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\-5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy............................

4, 5 tương tự nhé bn!

bài 3

1 (x-1)(x+2)+5x-5=0

=>(x-1)(x+2)+(5x-5)=o

=>(x-1)(x+2)+5(x-1)=0

=>(x-1)(x+2+5)=0

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

=>\(\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

vậy x=1 hoặc x=-7

2. (3x+5)(x-3)-6x-10=0

=>(3x+5)(x-3)-(6x+10)=0

=>(3x+5)(x-3)-2(3x+5)=0

=>(3x+5)(x-3-2)=0

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

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

Đáp án B

0,5 = 1/2

số lớn : 216 : (2+1) x 2 = 144

số bé : 216 - 144 = 72

Số bé là 72

Số lớn là 144

Ủng hộ mình đi

1: \(3x^2+mx+n\)

\(=3x^2+15x+\left(m-15\right)x+5m-75-5m+75+n\)

\(3x^2+mx+n\) chia x+5 dư 27

=>-5m+75+n=27

=>-5m+n=27-75=-48

2: \(x^3+mx+n\)

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

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

\(x^3+mx+n\) chia x+1 dư 7

=>n-m-1=7

=>n=m+1+7=m+8

\(x^3+mx+n\)

\(=x^3-3x^2+3x^2-9x+\left(m+9\right)x+3m+27+n-3m-27\)

\(=\left(x-3\right)\left(x^2+3x+m+9\right)+n-3m-27\)

\(x^3+mx+n\) chia x-3 dư 5

=>n-3m-27=5

=>n=3m+27+5=3m+32

mà n=m+8

nên 3m+32=m+8

=>2m=-24

=>m=-12

n=m+8=-12+8=-4

Đáp án A

Đáp án D

Yêu cầu đề bài là gì vậy bạn?

\(a,\Rightarrow 2x^2-10x-3x-2x^2=26\\ \Rightarrow -13x=26\\ \Rightarrow x=-2\\ b, \Rightarrow -2x^2+3x+3-3x-3+2x^2-x=18\\ \Rightarrow -x=18\Rightarrow x=-18\)

Lại là bạn cảm ơn

b: =x-2

d: \(=-x^3+\dfrac{3}{2}-2x\)