1.  \(6a^2-ab-15b^2=0\)

\(\Leftrightarrow6a^2-10ab+9ab-15b^2=0\)

\(\Leftrightarrow2a\left(3a-5b\right)+3b\left(3a-5b\right)=0\)

\(\Leftrightarrow\left(2a+3b\right)\left(3a-5b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=\frac{-3}{2}b\\a=\frac{5}{3}b\end{cases}}\)

-TH1:  \(a=\frac{-3}{2}b\)  thay vào M ta đc

\(M=\frac{11.\left(\frac{-3}{2}b\right)^2-2b.\frac{-3}{2}b+9b^2}{5\left(\frac{-3}{2}b\right)^2+3b.\frac{-3}{2}b+6b^2}=...\)

Tương tự cho TH2.

BÀi 3: b) Theo đề bài ta có Q(1) = 5; Q(14) = 9

Gọi số dư Q(x) chia cho (x-1)(x-14) là ax+b

=> Q(x) = P(x).(x-1)(x-14) + ax+b

Do đó Q(1) = P(x).(1-1)(1-14) + a.1 + b = a+b => a+b=5

và Q(14) = P(x).(14-1)(14-14) + a.14 + b = 14a+b => 14a+b=9

Giải hệ  \(\hept{\begin{cases}a+b=5\\14a+b=9\end{cases}}\)  tìm đc \(a=\frac{4}{13};b=\frac{61}{13}\)

Vậy số dư là  \(\frac{4}{13}x+\frac{61}{13}\)

Mong các bạn giúp mình, trong lúc hỏi mình sẽ luôn suy nghĩ chứ ko hoàn toàn dựa vào các bạn đâu, nếu bời ạn nào ra đáp án vui lòng ghi cả lời giải giúp mình

xin lỗi các bạn B2 là chia cho x²+2x dư -3x+2 nhé

Bạn tự giải luôn đi!

dài quá, ko muốn giải

Ta có:

\(a+b+c-abc=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)

\(=\left(a+b+c\right)\left(ab+c\left(a+b\right)\right)-abc\)

\(=\left(a+b\right)ab+\left(a+b\right)^2c+abc+c^2\left(a+b\right)-abc\)

\(=\left(a+b\right)\left(ab+c^2+c\left(a+b\right)\right)\)

\(=\left(a+b\right)\left(ab+ac+c^2+bc\right)\)

\(=\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)

\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

Đồng thời:

\(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)

Tương tự:

\(b^2+1=\left(a+b\right)\left(b+c\right)\)

\(c^2+1=\left(a+c\right)\left(b+c\right)\)

Từ đó:

\(P=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)}\)

\(=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}=1\)

\(N=\dfrac{\left(ab\right)^3+\left(bc\right)^3+\left(ca\right)^3}{\left(ab\right)\left(bc\right)\left(ca\right)}\)

Đặt \(\left(ab;bc;ca\right)=\left(x;y;z\right)\Rightarrow x+y+z=0\Rightarrow N=\dfrac{x^3+y^3+z^3}{xyz}\)

\(N=\dfrac{x^3+y^3+z^3-3xyz+3xyz}{xyz}=\dfrac{\dfrac{1}{2}\left(x+y+z\right)\left[\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\right]+3xyz}{xyz}=\dfrac{3xyz}{xyz}=3\)

\(\left(a+b+c\right)^2=a^2+b^2+c^2\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=a^2+b^2+c^2\)

\(\Leftrightarrow2\left(ab+bc+ac\right)=0\Leftrightarrow ab+bc+ac=0\Leftrightarrow bc=-ab-ac\)

\(\dfrac{a^2}{a^2+2bc}=\dfrac{a^2}{a^2+bc-ac-ab}=\dfrac{a^2}{\left(a-c\right)\left(a-b\right)}\)

CMTT: \(\left\{{}\begin{matrix}\dfrac{b^2}{b^2+2ca}=\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}\\\dfrac{c^2}{c^2+2ab}=\dfrac{c^2}{\left(c-a\right)\left(c-b\right)}=\dfrac{c^2}{\left(a-c\right)\left(b-c\right)}\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{a^2}{\left(a-c\right)\left(a-b\right)}+\dfrac{b^2}{\left(b-a\right)\left(b-c\right)}+\dfrac{c^2}{\left(a-c\right)\left(b-c\right)}=\dfrac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=1\)