ta có chu vi hcn(hình chữ nhật) là 44 nên nửa chu vi là 44:2=22

gọi x và y lần lượt là hcn (x;y<22 / đơn vị : cm)

ta có nửa chu vi là 22 nên : x+y=22 (1)

vì nếu tăng chiều dài lên 3cm , giảm chiều rộng 3cm thì S giảm \(21cm^2\) nên

(x+3).(y-3)=xy-21 (2)

theo (1) và (2) ta có :

\(\hept{\begin{cases}x+y=22\\\left(x+3\right).\left(y-3\right)=xy-21\end{cases}}\)=>\(\hept{\begin{cases}x+y=22\\xy-3x+3y-9=xy-21\end{cases}}\)

=>\(\hept{\begin{cases}x+y=22\\-3x+3y-9=-21\end{cases}}\)=>\(\hept{\begin{cases}x+y=22\\-3x+3y=-12\end{cases}}\)=>\(\hept{\begin{cases}x+y=22\\x-y=4\end{cases}}\)=>\(\hept{\begin{cases}x=13\\y=9\end{cases}}\)

        ta có chiều rộng là 9 ; chiều dài là 13

                                                                          vậy........

\(\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{5-2\sqrt{5}+1}\)

\(=\sqrt{\sqrt{5}^2-2\sqrt{5}+1^2}\)

\(=\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=|\sqrt{5}-1|=\sqrt{5}-1\)

bấm máy tính casio là ra đc đấy :))

b) ( 2x + 1 )² = 9

<=> ( 2x + 1 )² = 3²

<=> 2x + 1 = 3 hoặc 2x + 1 = -3

<=> 2x = 2 hoặc 2x = -4

<=> x = 1 hoặc x = -2

a) 9x² + 6x - 8 = 0

<=> 9x² + 12x -6x - 8 = 0

<=> 3x(3x+4) -2(3x+4) = 0

<=> (3x+4)(3x-2)=0

<=> \(\orbr{\begin{cases}3x+4=0\\3x-2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=-4\\3x=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{4}{3}\\x=\frac{2}{3}\end{cases}}}\)

Vậy ...

\(MN=4,8\)

\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\left(x\ne3;x\ne-1\right)\)

\(\Leftrightarrow\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\frac{2x\cdot2}{2\left(x-3\right)\left(x+1\right)}=0\)

\(\Leftrightarrow\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0\)

\(\Leftrightarrow\frac{2x^2-6x}{2\left(x-3\right)\left(x+1\right)}=0\)

\(\Leftrightarrow\frac{2x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=0\)

=> 2x=0

<=> x=0

Vậy x=0

+ Ta có: \(\frac{x}{2.\left(x-3\right)}+\frac{x}{2.\left(x+1\right)}=\frac{2x}{\left(x+1\right).\left(x-3\right)}\)\(\left(ĐKXĐ: x\ne-1, x\ne3\right)\)

      \(\Leftrightarrow\frac{x.\left(x+1\right)+x.\left(x-3\right)}{2.\left(x-3\right).\left(x+1\right)}=\frac{4x}{2.\left(x-3\right).\left(x+1\right)}\)

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

      \(\Leftrightarrow\left(x^2+x^2\right)+\left(x-3x-4x\right)=0\)

      \(\Leftrightarrow2x^2-6x=0\)

      \(\Leftrightarrow2x.\left(x-6\right)=0\)

      \(\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(TM\right)\\x=6\left(TM\right)\end{cases}}\)

Vậy \(S=\left\{0,6\right\}\)

+ Ta có: \(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^3-1}\)\(\left(ĐKXĐ:x\ne1,x^2+x+1\ne0\right)\)

       \(\Leftrightarrow\frac{\left(x^2+x+1\right)+2.\left(x-1\right)}{\left(x-1\right).\left(x^2+x+1\right)}=\frac{3x^2}{\left(x-1\right).\left(x^2+x+1\right)}\)

        \(\Rightarrow x^2+x+1+2x-2=3x^2\)

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

      \(\Leftrightarrow-2x^2+3x-1=0\)

      \(\Leftrightarrow2x^2-3x+1=0\)

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

      \(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)=0\)

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

      \(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=1\\x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(TM\right)\\x=1\left(L\right)\end{cases}}\)

Vậy \(S=\left\{\frac{1}{2}\right\}\)

Ko chép lại đề!

\(\Leftrightarrow x^2-7^2+x^2-2=2x^2+10\)

\(\Leftrightarrow x^2-49+x^2-2=2x^2+10\)

\(\Leftrightarrow2x^2-51=2x^2+10\)

<=> -51 = 10 ( vô lý )

=> \(x\in\varnothing\)

\(\left(x+7\right)\left(x-7\right)+x^2-2=2\left(x^2+5\right)\)

\(\Leftrightarrow x^2-49+x^2-2=2x^2+10\)

\(\Leftrightarrow x^2+x^2-2x^2=10+2+49\)

\(\Leftrightarrow0x=61\)(vô nghiệm)

\(\Leftrightarrow x\in\varnothing\)

1)(3x-2)(4x+3)=12x^2+x-6

2x(6x-1)=12x^2-2x

Suy ra 12x^2+x-6=12x^2-2x

Suy ra x-6=-2x

Suy ra x+2x=6

Suy ra 3x=6 Suy ra x=2

2)4x^2-(2x-1)(2x+1)=4x^2-(4x^2-1)=0

Suy ra 4x^2-4x^2+1=0

=> 1=0 =>Pt vô nghiệm

\(X^2-X+Y^2+Y+\frac{1}{2}=0\)

<=> \(\left(X^2-2X\frac{1}{2}+\frac{1}{4}\right)+\left(Y^2+2Y\frac{1}{2}+\frac{1}{4}\right)=0\)

<=>\(\left(X-\frac{1}{2}\right)^2+\left(Y+\frac{1}{2}\right)^2=0\)

Vì \(\left(X-\frac{1}{2}\right)^2\ge0\forall X\) ,   ,\(\left(Y+\frac{1}{2}\right)^2\ge0\forall Y\)

=> \(VT\ge0\forall X;Y\)

mà VT = 0

Từ 2 điều trên => \(\hept{\begin{cases}\left(X-\frac{1}{2}\right)^2=0\\\left(Y+\frac{1}{2}\right)^2=0\end{cases}}\)

<=>\(\hept{\begin{cases}X-\frac{1}{2}=0\\Y+\frac{1}{2}=0\end{cases}}\)

<=>\(\hept{\begin{cases}X=\frac{1}{2}\\Y=-\frac{1}{2}\end{cases}}\)

kết luận:

8) \(\left(x+4\right)\left(6x-12\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\6x-12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\6x=12\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-4\\x=2\end{cases}}}\)

Vậy \(x\in\left\{-4;2\right\}\)

11) \(\left(\frac{7}{8}-2x\right)\left(3x+\frac{1}{3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{7}{8}-2x=0\\3x+\frac{1}{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=\frac{7}{8}-0\\3x=-\frac{1}{3}\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=\frac{7}{8}\\x=-\frac{1}{9}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{16}\\x=-\frac{1}{9}\end{cases}}}\)

Vậy \(x\in\left\{\frac{7}{16};-\frac{1}{9}\right\}\)

12) \(3x-2x^2=0\)

\(\Leftrightarrow x\left(3-2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)

Vậy \(x\in\left\{0;\frac{3}{2}\right\}\)

13) \(5x+10x^2=0\)

\(\Leftrightarrow5x\left(1+2x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{1}{2}\end{cases}}\)

Vậy \(x\in\left\{0;-\frac{1}{2}\right\}\)