\(\frac{x}{4}=\frac{y}{7}\) và xy=112

Đặt \(\frac{x}{4}=\frac{y}{7}=k\Rightarrow x=4k;y=7k\)

Mà xy=112 hay 4k . 7k=112

                  <=> 28k^2 =112

                   <=> k^2 =4

                    <=>k=2 ; k=-2

Với k=2 thì x=8 ; y=14

Với k=-2 thì x=-8 ; y=-14

a) \(x\left(x-1\right)-x^2+2x=5\)

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

\(\Leftrightarrow x=5\)

b) \(4x\left(3x+2\right)-6x\left(2x+5\right)+21\left(x-1\right)=0\)

\(\Leftrightarrow12x^2+8x-12x^2-30x+21x-21=0\)

\(\Leftrightarrow-x=21\)

\(\Leftrightarrow x=-21\)

c) \(2x\left(x+1\right)-x^2\left(x+2\right)+x^3-x+4=0\)

\(\Leftrightarrow2x^2+2x-x^3-2x^2+x^3-x+4=0\)

\(\Leftrightarrow x=-4\)

Bài 1:
a) Để A,B có nghĩa \(\Leftrightarrow\begin{cases}2x+3\ge0\\x-3>0\end{cases}\)\(\Leftrightarrow\begin{cases}x\ge-\frac{3}{2}\\x>3\end{cases}\)\(\Leftrightarrow x>3\)

b) Để A= B

\(\Leftrightarrow\sqrt{\frac{2x+3}{x-3}}=\frac{\sqrt{2x+3}}{\sqrt{x-3}}\)

\(\Leftrightarrow\sqrt{\frac{2x+3}{x-3}}-\sqrt{\frac{2x+3}{x-3}}=0\)

\(\Leftrightarrow0x=0\) (thỏa mãn với mọi x>3)

Vậy x>3 thì A=B

\(A=\frac{3}{n+1}\) 

Để A nguyên thì n+1\(\in\)Ư(3)

Mà Ư(3)={1;-1;3;-3}

Ta có bảnh sau:

Vậy x={-4;-2;0;2}

\(B=\frac{3n+5}{n+4}=\frac{3\left(n+4\right)-7}{n+4}=3-\frac{7}{n+4}\)

Vậy để B nguyên thì n+4 thuộc Ư{7}

Mà:Ư(7)={1;-1;7;-7}

=>n+4={1;-1;7;-7}

Ta có bẳng sao:

VaVaayk x={-11;-5;-3;3}

a) \(8\left(x-2\right)-2\left(3x-4\right)=2\)

\(\Leftrightarrow8x-16-6x+8=2\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=5\)

b) Xem lại đề

co ghi lon de bai ko? dien tich = 60 la sai

a) \(3x\left(x-4y\right)-\frac{12}{5}y\left(y-5x\right)=3x^2-12xy-\frac{12}{5}y^2+12xy=3x^2-\frac{12}{5}y^2\)

b) \(\left(4x^2-3y\right)\cdot2y-\left(3x^2-4y\right)\cdot3y\)

\(=8x^2y-6y^2-9x^2y+12y^2=-x^2y+6y^2\)

a) \(\sqrt{\left(x-3\right)^2}=9\)

\(\Leftrightarrow\left|x-3\right|=9\)   (1)

+)TH1: \(x\ge3\) thì pt (1) trở thành:

\(x-3=9\Leftrightarrow x=12\left(TM\right)\)

+)TH2: \(x< 3\) thì pt (1) trở thành:

\(3-x=9\Leftrightarrow x=-6\left(TM\right)\)

Vậy x={-6;12}

b) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow2x+1=6\)

\(\Leftrightarrow x=\frac{5}{2}\)

a) \(\sqrt{2}x-\sqrt{50}=0\)

\(\Leftrightarrow\sqrt{2}x=\sqrt{50}\)

\(\Leftrightarrow x=\frac{\sqrt{50}}{\sqrt{2}}=\sqrt{\frac{50}{2}}=\sqrt{25}=5\)

b) \(\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}\)

\(\Leftrightarrow\sqrt{3}\left(x+1\right)=2\sqrt{3}+3\sqrt{3}\)

\(\Leftrightarrow\sqrt{3}x=5\sqrt{3}\)

\(\Leftrightarrow x=5\)

c) \(\sqrt{3}x^2-\sqrt{12}=0\)

\(\Leftrightarrow\sqrt{3}\left(x^2-2\right)=0\)

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

\(\Leftrightarrow x^2=2\Leftrightarrow\left[\begin{array}{nghiempt}x=\sqrt{2}\\x=-\sqrt{2}\end{array}\right.\)

d) \(\frac{x^2}{\sqrt{5}}-\sqrt{20}=0\)

\(\Leftrightarrow\)\(\frac{1}{\sqrt{5}}\left(x^2-10\right)=0\)

\(\Leftrightarrow x^2-10=0\)

\(\Leftrightarrow x^2=10\Leftrightarrow\left[\begin{array}{nghiempt}x=\sqrt{10}\\x=-\sqrt{10}\end{array}\right.\)

Đề j mà kì quậc thế này