cạnh là 100+10=110%

diện tick hình vuông mới là 110*110/100=121%

diện tích hình cũ là 100*100/100=100%

tăng số phần trăm là 121-100=21% 

tick nhé

Bài 1:

\(x^2-5x-6=0\)

\(\Leftrightarrow x^2+x-6x-6=0\)

\(\Leftrightarrow\left(x^2+x\right)-\left(6x+6\right)=0\)

\(\Leftrightarrow x\left(x+1\right)-6\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=6\end{matrix}\right.\)

Vậy x=-1; x=6

Bài 2:

a) Ta có: \(x+y=10\Leftrightarrow y=10-x\) (1)

Từ (1) thay vào \(P=xy\) ta được:

\(P=x\left(10-x\right)\)

\(\Leftrightarrow P=10x-x^2\)

\(\Leftrightarrow P=-x^2+10x-5^2+5^2\)

\(\Leftrightarrow P=-\left(x^2-10x+5^2\right)+5^2\)

\(\Leftrightarrow P=-\left(x-5\right)^2+25\)

Vậy GTLN của P=25 khi \(x-5=0\Leftrightarrow x=5\)

b) \(P=x^2-5x\)

\(\Leftrightarrow P=x^2-2x.\dfrac{5}{2}+\left(\dfrac{5}{2}\right)^2-\left(\dfrac{5}{2}\right)^2\)

\(\Leftrightarrow P=\left(x-\dfrac{5}{2}\right)^2-\dfrac{25}{4}\)

Vậy GTNN của \(P=\dfrac{-25}{4}\) khi \(x-\dfrac{5}{2}=0\Leftrightarrow x=\dfrac{5}{2}\)

1)\(Q=-x^2+2x-7\)

\(\Leftrightarrow Q=-x^2+2x-1-6\)

\(\Leftrightarrow Q=-\left(x^2-2x+1\right)-6\)

\(\Leftrightarrow Q=-\left(x-1\right)^2-6\)

Vậy GTLN của \(Q=-6\) khi \(x-1=0\Leftrightarrow x=1\)

2)\(\left(x-5\right)\left(2x+3\right)-2x\left(x-3\right)+x-7\)

\(=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x-7\)

\(=2x^2+3x-10x-15-2x^2+6x+x-7\)

\(=-15-7\)

\(=-22\)

\(\left(2x+3\right)^2+\left(2x-3\right)^2-\left(2x+3\right)\left(4x-6\right)+xy\)

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

\(=\left[\left(2x+3\right)^2-2\left(2x+3\right)\left(2x-3\right)+\left(2x-3\right)^2\right]+xy\)

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

\(=6^2+xy\)

\(=36+xy\)

1) \(3x^2-6xy+3y^2-12z^2\)

\(=3\left(x^2-2xy+y^2-4z^2\right)\)

\(=3\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

\(=3\left(x-y-2z\right)\left(x-y+2z\right)\)

2)

a) \(3\left(x-1\right)^2-3x\left(x-5\right)-2=0\)

\(\Leftrightarrow3\left(x^2-2x+1\right)-3x\left(x-5\right)-2=0\)

\(\Leftrightarrow3x^2-6x+3-3x^2+15x-2=0\)

\(\Leftrightarrow9x+1=0\)

\(\Leftrightarrow9x=-1\)

\(\Leftrightarrow x=\dfrac{-1}{9}\)

Vậy \(x=\dfrac{-1}{9}\)

b) \(2x^2-5x-7=0\)

\(\Leftrightarrow2x^2+2x-7x-7=0\)

\(\Leftrightarrow\left(2x^2+2x\right)-\left(7x+7\right)=0\)

\(\Leftrightarrow2x\left(x+1\right)-7\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\2x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\2x=7\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{2}\end{matrix}\right.\)

Vậy \(x=-1\); \(x=\dfrac{7}{2}\)

đề 1

1) A

2) A

3) C

4) A

5) A

6) B

Cau 2

sai

dung

dung

dung

7 và 55

5 và 77

\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4x^2\)

\(\Leftrightarrow P=\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2-4x^2\)

\(\Leftrightarrow P=\left(x+y-x+y\right)^2-4x^2\)

\(\Leftrightarrow P=\left(2y\right)^2-4x^2\)

\(\Leftrightarrow P=4y^2-4x^2\)