\(3x\left(x-y\right)+x-y\)

\(=3x\left(x-y\right)+1\left(x-y\right)\)

\(=\left(x-y\right)\left(3x+1\right)\)

\(=\left(x-y\right)\left(3x+1\right)\)

\(a,=15x^4-12x^3+9x^2\\ b,=-15x^3y^2+25x^2y^2-5xy^3\\ c,=5x^3-19x^2+12x\\ d,=3x^3+xy^2+5x^2y-9x^2y-3y^3-15xy^2\\ =3x^3-3y^3-14xy^2-4x^2y\)

mình cảm ơn

vs đk tổng =1 ta có:

\(\dfrac{a+bc}{b+c}+\dfrac{b+ca}{c+a}+\dfrac{c+ab}{a+b}\)

\(=\dfrac{a\left(a+b+c\right)+bc}{bc}+\dfrac{b\left(a+b+c\right)+ca}{ca}+\dfrac{c\left(a+b+c\right)+ab}{ab}\)

\(=\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(b+c\right)\left(b+a\right)}{c+a}+\dfrac{\left(c+a\right)\left(c+b\right)}{a+b}\)

sd bđt AM-GM cho 2 số dương ta có:

\(\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(b+c\right)\left(b+a\right)}{c+a}\ge2\left(a+b\right)\)

\(\dfrac{\left(b+c\right)\left(b+a\right)}{c+a}+\dfrac{\left(c+a\right)\left(c+b\right)}{a+b}\ge2\left(b+c\right)\)

\(\dfrac{\left(a+b\right)\left(a+c\right)}{b+c}+\dfrac{\left(c+a\right)\left(c+b\right)}{a+b}\ge2\left(c+a\right)\)

Cộng theo vế 3 đẳng thức trên ta sẽ có điều phải chứng minh 

Đẳng thức xảy ra khi và chỉ khi a = b= c =\(\dfrac{1}{3}\)

Không biết nãy bị lỗi ở đâu, mình gửi lại:<

1 does this red dress cost

2 is this shirt

does this short cost

3 do these shoes cost

4 has brown backpack

5 how to get to the bus stop

6 to stay in bed

7 are cleaned by Nga everyday

8 is played with a bow

9 worked for that company for 10 years

10 15 minutes walking to school

23:

 u4=10 và u7=22

=>\(\left\{{}\begin{matrix}u1+3d=10\\u1+6d=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3d=-12\\u1+3d=10\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}d=4\\u1=10-12=-2\end{matrix}\right.\)

=>Chọn C

Câu 22:

\(\left\{{}\begin{matrix}u1+2u5=0\\S_4=14\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}u1+2\left(u1+4d\right)=0\\4\cdot\dfrac{\left[2u1+3d\right]}{2}=14\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3u1+8d=0\\2u1+3d=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6u1+16d=0\\6u1+9d=21\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7d=-21\\2u_1+3d=7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}d=-3\\2u_1=7-3d=7+9=16\end{matrix}\right.\)

=>\(u_1=8;d=-3\)

=>Chọn A

21A

19B

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

b) \(=\left(x^2-6x+9\right)-y^2=\left(x-3\right)^2-y^2=\left(x-y-3\right)\left(x+y-3\right)\)

c) \(=5\left(x^2-xy-2x+2y\right)=5\left[x\left(x-y\right)-2\left(x-y\right)\right]=5\left(x-y\right)\left(x-2\right)\)

đầu bài là phân tích đa thức thành phân tử nha mn

mình cũng ko bt á

\(⇔ 5x+3=4x+12\)

\(⇔ x =9\)