a: =>x=13/52+8/52=21/52

b: =>x=1/36-27/36=-26/36=-13/18

c: =>x=24/60+15/60-20/60=19/60

d: =>x/15=9/15-10/15=-1/15

=>x=-1

a, x = 21/52 ; b x = -13/18 

c, x = 19/60 ; d, x = -1 

tui mới lớp 4

a, ta có A(x)=2x³+7x²+ax+b

                   =(2x³+2x²+2x)+(5x²+5x+5)+ax-7x+b-5

                   =2x(x²+x+1)+5(x²+x+1)+(a-7)x+(b-5)

                   =(x²+x+1)(2x+5)+(a-7)x+(b-5)

ta có: (x²+x+1)(2x+5)⋮B(x)

→để A(x)⋮B(x) thì (a-7)x+(b-5)=0

→\(\left\{{}\begin{matrix}a-7=0\\b-5=0\end{matrix}\right.\) ⇔\(\left\{{}\begin{matrix}a=7\\b=5\end{matrix}\right.\)

vậy ....

mk trình bày hơi tắt xíu 

bn cố gắng dịch nhé

a) \(A\left(x\right)=2x^3+7x^2+ax+b=\left(x^2+x-1\right)\left(2x+5\right)+\left(a-3\right)x+b+5\)

\(=\left(2x+5\right)B\left(x\right)+\left(a-3\right)x+b+5\)

Để \(A\left(x\right)\) chia hết cho \(B\left(x\right)\) thì 

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

\(\Leftrightarrow\left\{{}\begin{matrix}a-3=0\\b+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=-5\end{matrix}\right.\)

b) \(A\left(x\right)=ax^3+bx-24=\left(ax-4a\right)\left(x^2+4x+3\right)+\left(b+13a\right)x+12a-24\)

\(=\left(ax-4a\right)\left(x^2+4x+3\right)+\left(b+13a\right)x+12a-24\)

Để \(A\left(x\right)\) chia hết cho \(B\left(x\right)\) thì 

\(\left(b+13a\right)x+12a-24=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}b+13a=0\\12a-24=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=2\\b=-26\end{matrix}\right.\)

c) Tương tự. 

1. you be quiet, you'll hear the sound.

2. I am short, I can't become a model.

3. you don't run quickly, you can't catch the bus.

4. Jack doesn't pass the exam, his parents will be sad.

5. you don't be hurry, you'll be late.

6. they don't tell the truth, we can't know about this.

7. you speak slowly, I'll understand you.

Từ giả thiết, suy ra: \(\hat{M}=\dfrac{3}{2}\hat{P}\).

Ta có: \(\hat{D}+\hat{M}+\hat{P}=180^o\) (tổng 3 góc trong một tam giác)

\(\Leftrightarrow55^o+\dfrac{3}{2}\hat{P}+\hat{P}=180^o\Leftrightarrow\hat{P}=50^o\)

\(\Rightarrow\hat{M}=\dfrac{3}{2}\hat{P}=\dfrac{3}{2}\cdot50^o=75^o\)

vcl