\(a-b=9\Rightarrow\left(a-b\right)^2=a^2-2ab+b^2=81.\)

\(\Rightarrow a^2+b^2=81+2ab=81+2.36=153\)

\(P=\left(a+b\right)^2=a^2+b^2+2ab=153+2.36=225\)

Answer:

a) Theo đề ra:

\(a^2+2c^2=3b^2+19\)

\(\Rightarrow a^2+2c^2-3b^2=19\)

Có:

\(\frac{a^2+7}{4}=\frac{b^2+6}{5}=\frac{c^2+3}{6}=\frac{3b^2+18}{15}=\frac{2c^2+6}{12}=\frac{a^2+7+2c^2+6-3b^2-18}{4+12-15}=14\)

\(\Rightarrow\hept{\begin{cases}a^2=49\Rightarrow a=7\\b^2=64\Rightarrow b=8\\c^2=81\Rightarrow c=9\end{cases}}\)

b) \(P=x^4+2x^3+3x^2+2x+1\)

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

\(=\left(x^2+1\right)^2+2x\left(x^2+1\right)+x^2\)

\(=\left(x^2+x+1\right)^2\)

Mà:

 \(x^2+x+1\)

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

\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\) nên \(P\ge\left(\frac{3}{4}\right)^2=\frac{9}{16}\)

Dấu "=" xảy ra khi: \(x=\frac{-1}{2}\)

7 + 4 = 11 nha

~HT~

7 + 4 = 11 chứ nhiu. cái câu này đến lớp 1 cũng biết hỏi. hỏi linh tinh

Answer:

Bài 1:

\(x^2-xy+2x-2y\)

\(=x\left(x-y\right)+2\left(x-y\right)\)

\(=\left(x-y\right)\left(x+2\right)\)

\(x^2-4y^2\)

\(=\left(x\right)^2-\left(2y\right)^2\)

\(=\left(x-2y\right)\left(x+2y\right)\)

Bài 2:

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

\(\Rightarrow5\left(x-3\right)+x\left(x-3\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-5\end{cases}}}\)

\(x^2-x-12=0\)

\(\Rightarrow x^2+3x-4x-12=0\)

\(\Rightarrow x\left(x+3\right)-4\left(x+3\right)=0\)

\(\Rightarrow\left(x+3\right)\left(x-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}}\)

Bài 3:

a, Với \(x=-3\)

\(A=\frac{x-4}{x+5}=\frac{-3-4}{-3+5}=-\frac{7}{2}\)

b, \(B=\frac{2}{x+4}+\frac{x+20}{x^2-16}\left(ĐK:x\ne\pm4;x\ne-5\right)\)

\(=\frac{2}{x+4}+\frac{x+20}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{2\left(x-4\right)+x+20}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{2x-8+x+20}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{3x+12}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{3\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}\)

\(=\frac{3}{x-4}\)

c, \(M=A.B=\frac{x-4}{x+5}.\frac{3}{x-4}=\frac{3}{x+5}\)

Để M nguyên thì \(3⋮\left(x+5\right)\)

\(\Rightarrow x+5\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow x\in\left\{-4;-6;-2;-8\right\}\) mà \(ĐK:x\ne\pm4\)

Vậy \(x\in\left\{-6;-2;-8\right\}\)

D. 4 nhé

D. 4 nhaa!! 

16x³+54y³

= 16.(x³+4y³)

\(16x^3+54y^3\)

\(=2\left(8x^3-27y^3\right)\)

\(=2.\left(2x-3y\right)\left(4x^2-6xy+9y^2\right)\)

(Không ghi đề bài)

= (x - 5)²/ [x.(x² - 25)]

=  (x - 5)² / [x.(x - 5).(x+5)]

= (x-5) / x.(x+5)

Thay x = -3/5

=> (-3/5-5) / [-3/5.(-3/5+5)]

= -28/5 : (-3/5 . 22/5)

= -28/5 : (-66/25)

=  -28/5 . -25/66

= 70/33

Đây nhé!! Chúc bạn học tốt!!✨