1) -12.(x-5) + 7.(3-x)=5

   -12x+ 60+21-7x =5

    -12x-7x = 5-60-21

    -19x=-76

     x=-76:(-19)

     x=4

2) (x-2).(x+4) =0

   \(\Rightarrow\)x-2=0 hoặc x+4=0

x-2=0                    x+4=0

x=0+2                    x=0-4     

x=2                         x=-4

Vậy x=2 hoặc x=-4

3) (x-2).(x+15) =0

 \(\Rightarrow\)x-2=0 hoặc x+15=0

 x-2=0                  x+15=0

x=0+2                   x=0-15 

 x=2                       x=-15

1)\(-12.\left(x-5\right)+7.\cdot\left(3-x\right)=5\)

\(-12x+60+21-7x=5\)

\(-19x+81=5\)

\(-19x=5-81\)

-\(-19x=-76\)

\(x=-76:-19\)

\(x=4\)

2) Ta có 2 trường hợp

TH1: x-2=0 =>x=2

TH2: x+4=0 => x=-4

Vậy \(x\in\left(-4;2\right)\)

3) Ta có

TH1: x-2=0=>x=2

TH2: x+15=0=>x=-15

Vậy \(x\in\left(-15;2\right)\)

2)   (x - 2).(x + 4) = 0

\(\hept{\begin{cases}x-2=0\\x+4=0\end{cases}\Rightarrow\hept{\begin{cases}x=0+2\\x=0-4\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\x=-4\end{cases}}}\)

Vậy x = 2 và (-4)

3)   (x-2).(x + 15) = 0

\(\hept{\begin{cases}x-2=0\\x+15=0\end{cases}\Rightarrow}\hept{\begin{cases}x=0+2\\x=0-15\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\x=-15\end{cases}}\)

Vậy x = 2 và (-15)

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

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

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

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

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

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

Vậy \(S=\left\{5;-2\right\}\)

\(\Leftrightarrow\left(x-5\right)^2+7\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x-5+7\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Câu 1:

\(2x^3-3x^2+x+a\)

\(=2\left(x^3-6x^2+12x-8\right)+9\left(x^2-4x+4\right)+13\left(x-2\right)+\left(6+a\right)\)

\(=2\left(x-2\right)^3+9\left(x-2\right)^2+13\left(x-2\right)+\left(6+a\right)\)chia hết cho \(x-2\)khi và chỉ khi :

\(6+a=0\Leftrightarrow a=-6\). Vậy \(a=-6\).

Câu 2:

\(\left(x+1\right)\left(2x-x\right)-\left(3x+5\right)\left(x+2\right)=4x^2+1\)

\(\Leftrightarrow x^2+x-\left(3x^2+11x+10\right)=-4x^2+1\)

\(\Leftrightarrow x^2+x-3x^2-11x-10+4x^2-1=0\)

\(\Leftrightarrow2x^2-10x-11=0\)

\(\Delta'=\left(-5\right)^2-2\left(-11\right)=47>0\)

\(\Rightarrow\)Phương trình có 2 nghiệm phân biệt:

\(x=\frac{5+\sqrt{47}}{2}\)hoặc \(x=\frac{5-\sqrt{47}}{2}\)

Vậy phương trình có tập nghiệm \(S=\left\{\frac{5+\sqrt{47}}{2};\frac{5-\sqrt{47}}{2}\right\}\)

a)( x- 1)( x- 5)= 0.

=> \(\orbr{\begin{cases}x-1=0\\x-5=0\end{cases}}\).

=> \(\orbr{\begin{cases}x=1\\x=5\end{cases}}\).

Vậy x= 1 hoặc x= 5 thì( x- 1)( x- 5)= 0.

\(\left(x-1\right)\left(x-5\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-5=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0+1\\x=0+5\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=1\\x=5\end{cases}}\)

Vậy x = 1 hoặc x = 5

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

\(\Rightarrow\orbr{\begin{cases}x+6=0\\x^2-1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0-6\\x^2=0+1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-6\\x^2=1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-6\\x=1\end{cases}}\)

Vậy x = -6 hoặc x = 1

\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\)

\(\Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\dfrac{x+200}{5}+\dfrac{20}{5}-4=0\)

\(\Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\)

\(\Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\)

\(\Leftrightarrow x=-200\)( do \(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}>0\))

\(\dfrac{x+1}{199}+\dfrac{x+2}{198}+\dfrac{x+3}{197}+\dfrac{x+4}{196}+\dfrac{x+220}{5}=0\\ \Leftrightarrow\left(\dfrac{x+1}{199}+1\right)+\left(\dfrac{x+2}{198}+1\right)+\left(\dfrac{x+3}{197}+1\right)+\left(\dfrac{x+4}{196}+1\right)+\left(\dfrac{x+220}{5}-4\right)=0\\ \Leftrightarrow\dfrac{x+200}{199}+\dfrac{x+200}{198}+\dfrac{x+200}{197}+\dfrac{x+200}{196}+\dfrac{x+200}{5}=0\\ \Leftrightarrow\left(x+200\right)\left(\dfrac{1}{199}+\dfrac{1}{198}+\dfrac{1}{197}+\dfrac{1}{196}+\dfrac{1}{5}\right)=0\\ \Leftrightarrow x=-200\)

a) \(...\Rightarrow\left\{{}\begin{matrix}x-2=0\\y+3=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=-3\end{matrix}\right.\)

b) \(...\Rightarrow|x-2|=|x+3|\Rightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-x-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}0x=5\\2x=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x=-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow x=-\dfrac{1}{2}\)

c) \(|x-\dfrac{3}{4}|+|x+\dfrac{5}{4}|=1\)

\(\Rightarrow\left\{{}\begin{matrix}x-\dfrac{3}{4}\le0\\x+\dfrac{5}{4}\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\le\dfrac{3}{4}\\x\ge-\dfrac{5}{4}\end{matrix}\right.\)

\(\Rightarrow-\dfrac{5}{4}\le x\le\dfrac{3}{4}\)

x-5/2<0

=>x>5/2

7-x/3>

=>x/3<7

(14,78-a)/(2,87+a)=4/1

14,78+2,87=17,65

Tổng số phần bằng nhau là 4+1=5

Mỗi phần có giá trị bằng 17,65/5=3,53

=>2,87+a=3,53

=>a=0,66.

bexiu đang trả lời câu nào vậy ghi giúp mik cái đề