a, {2x - 11y = -7
    {10x + 11y = 31
<=>{10x-55y=-35
       {10x+11y=31
<=>{-66y=-66
      {2x-11y=-7
<=>{2x-11.1=-7
       {y=1
<=>{x=2

\(\hept{\begin{cases}2x-11y=-7\\10x+11y=31\end{cases}}< =>\hept{\begin{cases}12x=24\\10x+11y=31\end{cases}}< =>\hept{\begin{cases}x=2\\y=1\end{cases}}\)

a, A=15√x−11x+2√x−3+3√x−21−√x−2√x+3√x+3A=15x−11x+2x−3+3x−21−x−2x+3x+3

=15√x−11x−√x+3√x−3−3√x−2√x−1−2√x+3√x+3=15x−11x−x+3x−3−3x−2x−1−2x+3x+3

=15√x−11√x(√x−1)+3(√x−1)−3√x−2√x−1−2√x+3√x+3=15x−11x(x−1)+3(x−1)−3x−2x−1−2x+3x+3

=15√x−11(√x−1)(√x+3)−3√x−2√x−1−2√x+3√x+3=15x−11(x−1)(x+3)−3x−2x−1−2x+3x+3

=15√x−11−(3√x−2)(√x+3)−(2√x+3)(√x−1)(√x−1)(√x+3)=15x−11−(3x−2)(x+3)−(2x+3)(x−1)(x−1)(x+3)

=15√x−11−(3x+9√x−2√x−6)−(2x−2√x+3√x−3)(√x−1)(√x+3)=15x−11−(3x+9x−2x−6)−(2x−2x+3x−3)(x−1)(x+3)

=15√x−11−3x−9√x+2√x+6−2x+2√x−3√x+3(√x−1)(√x+3)=15x−11−3x−9x+2x+6−2x+2x−3x+3(x−1)(x+3)

=7√x−5x−8(√x−1)(√x+3)

\(P=\frac{a^2+3}{b+c}+\frac{b^2+3}{c+a}+\frac{c^2+3}{a+b}\)

\(=\left(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\right)+3\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\)

\(\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}+3.\frac{9}{2\left(a+b+c\right)}\)

\(=\frac{3}{2}+\frac{9}{2}=6\)

Dấu \(=\)khi \(a=b=c=1\).

a, \(P=\frac{a^3-a+2b-\frac{b^2}{a}}{\left(1-\sqrt{\frac{a+b}{a^2}}\right)\left(a+\sqrt{a+b}\right)}:\left[\frac{a^2\left(a+b\right)+a\left(a+b\right)}{\left(a-b\right)\left(a+b\right)}+\frac{b}{a-b}\right]\)

\(=\frac{\frac{a^4-a^2-2ab-b^2}{a}}{\frac{\left(a-\sqrt{a+b}\right)\left(a+\sqrt{a+b}\right)}{a}}:\left[\frac{\left(a+b\right)\left(a^2+a\right)}{\left(a+b\right)\left(a-b\right)}+\frac{b}{a-b}\right]\)

\(=\frac{a^4-a^2-2ab-b^2}{a^2-a-b}:\frac{a^2+a+b}{a-b}\)

\(=\frac{a^4-a^2-2ab-b^2}{a^2-\left(a+b\right)}.\frac{a-b}{a^2+\left(a+b\right)}\)

\(=\frac{\left(a^4-a^2-2ab-b^2\right).\left(a-b\right)}{a^4-\left(a+b\right)^2}=\frac{\left[a^4-\left(a+b\right)^2\right].\left(a-b\right)}{a^4-\left(a+b\right)^2}=a-b\)

b, Có \(P=a-b=1\)\(\Rightarrow a=1+b\)

\(a^3-b^3=7\Leftrightarrow\left(a^2+ab+b^2\right)\left(a-b\right)=7\)

\(\Rightarrow a^2+ab+b^2=7\)

\(\Leftrightarrow\left(1+b\right)^2+\left(1+b\right)b+b^2=7\)

\(\Leftrightarrow b^2+2b+1+b^2+b+b^2=7\)

\(\Leftrightarrow3b^2+3b-6=0\)

Bạn tự giải phương trình tìm b => a

Bài 2 :

\(a,y=\left(m+1\right)x-2m-5\) \(\Leftrightarrow\left(m+1\right)x-2m-5-y=0\)

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

Có y luôn qua điểm A cố định với A( x₀ ; y₀ ) \(\orbr{\begin{cases}x_0-2=0\\x_0-y_0-5=0\end{cases}}\Rightarrow\orbr{\begin{cases}x_0=2\\y_0=-3\end{cases}}\)

=> A( 2;-3)

Gọi H là chân đường vuông góc hạ từ O xuống d => \(OH\le OA\)

\(OH_{max}=OA\)khi \(H\equiv A\)\(\left(d\perp OA\right)\)

=> đường thẳng OA qua O( 0;0 ) và A( 2;-3 ) => \(y=-\frac{3}{2}x\)

\(\Rightarrow d\perp OA\)=> hệ số góc \(m.\) \(-\frac{3}{2}=-1\Rightarrow m=\frac{2}{3}\)

b, \(y=0\Rightarrow\left(m+1\right)x-2m-5=0\)\(\Rightarrow x=\frac{2m+5}{m+1}\)\(\Rightarrow A\left(\frac{2m+5}{m+1};0\right)\)

\(x=0\Rightarrow y=-2m-5\Rightarrow B\left(0;-2m-5\right)\)

\(\Rightarrow OA=\sqrt{\frac{2m+5}{m+1}};OB=\sqrt{-2m-5}\)

\(\Rightarrow\frac{1}{2}.OA.OB=\frac{3}{2}\Rightarrow OA.OB=3\)

\(\Rightarrow\left(OA.OB\right)^2=9\Rightarrow\frac{\left(2m+5\right)^2}{m+1}=9\)

\(\Rightarrow4m^2+20m+25-9m-9=\)

\(\Rightarrow4m^2+11m+16=0\)

Đáp án: 123444444

^ HT ^

123444444

Gọi giao điểm của 2 đường thẳng đó trên trục tung là A( 0;a )

Khi đó tọa độ điểm A( 0;a ) thỏa mãn hpt \(\hept{\begin{cases}a=m^2+1\\a=5\end{cases}}\)

\(\Rightarrow m^2+1=5\)

\(\Rightarrow m^2=4\)

\(\Rightarrow m=\pm2\)

Vậy \(m=\pm2\)

chịu bạn ơi

chịu

bạn làm như thế thì toang mik rồi

Cách 1 :

\(\hept{\begin{cases}3x-2y=9\\x-3y=10\end{cases}\Leftrightarrow}\hept{\begin{cases}3x-2y=9\left(1\right)\\3x-9y=30\left(2\right)\end{cases}}\)

Trừ (1) và (2) ta được :

\(3x-2y-\left(3x-9y\right)=9-30\)

\(\Leftrightarrow3x-2y-3x+9y=-21\)

\(\Leftrightarrow7y=-21\Leftrightarrow y=-3\left(3\right)\)

Thay \(\left(3\right)\)vào \(\left(1\right)\) ta được :

\(3x-2y=9\Leftrightarrow3x-2\left(-3\right)=9\)

\(\Leftrightarrow3x+6=9\)

\(\Leftrightarrow3x=3\Leftrightarrow x=1\)

Cách 2 :

\(\hept{\begin{cases}3x-2y=9\\x-3y=10\end{cases}}\Leftrightarrow\hept{\begin{cases}2y=3x-9\\x-3y=10\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=\frac{3x-9}{2}\\x-3\cdot\frac{3x-9}{2}=10\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=\frac{3x-9}{2}\\x-\frac{9x-27}{2}=10\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=\frac{3x-9}{2}\\\frac{2x}{2}-\frac{9x-27}{2}=\frac{20}{2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=\frac{3x-9}{2}\\2x-9x+27=20\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=\frac{3x-9}{2}\\-7x=-7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}y=\frac{3\cdot1-9}{2}=-3\\x=1\end{cases}}\)

Vậy \(\left(x,y\right)=\left(1,-3\right)\)

C1 . Hệ phương trình \(\hept{\begin{cases}x=10+3y\\3\left(10+3y\right)-2y=9\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=10+3y\\30+9y-2y=9\end{cases}}\Leftrightarrow\hept{\begin{cases}x=10+3y\\y=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)

C2 : Hệ phương trình 

\(\hept{\begin{cases}x=3+\frac{2}{3}y\\3+\frac{2}{3}y-3y=10\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3+\frac{2}{3}y\\y=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x=1\\y=-3\end{cases}}\)