\(\dfrac{3}{x}+\dfrac{6}{y}=\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{6}{2x}+\dfrac{6}{y}=\dfrac{1}{4}\)

\(\Leftrightarrow6\left(\dfrac{1}{2x}+\dfrac{1}{y}\right)=\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{1}{2x}+\dfrac{1}{y}=\dfrac{1}{24}^{\left(1\right)}\)

Lại có: \(\dfrac{1}{x}+\dfrac{1}{y}=\dfrac{1}{16}^{\left(2\right)}\)

Lấy ⁽²⁾ trừ ⁽¹⁾ ta có:

\(\dfrac{1}{x}+\dfrac{1}{y}-\dfrac{1}{2x}-\dfrac{1}{y}=\dfrac{1}{16}-\dfrac{1}{24}\)

\(\Leftrightarrow\dfrac{2-1}{2x}=\dfrac{1}{48}\)

\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{48}\)

=> 2x = 48

<=> x = 24

Thay x = 24 vào ⁽²⁾ ta có:

\(\dfrac{1}{24}+\dfrac{1}{y}=\dfrac{1}{16}\)

\(\Leftrightarrow\dfrac{1}{y}=\dfrac{1}{48}\)

=> y = 48

Vậy ...

Ta có: \(\dfrac{3}{x}\) + \(\dfrac{6}{y}\) = \(\dfrac{1}{4}\)

<=> 3(\(\dfrac{1}{x}\) + \(\dfrac{2}{y}\) ) = \(\dfrac{1}{4}\)

<=> \(\dfrac{1}{x}\) + \(\dfrac{2}{y}\) = \(\dfrac{1}{12}\) (1)

Mặt khác: \(\dfrac{1}{x}\) + \(\dfrac{1}{y}\) = \(\dfrac{1}{16}\) (2)

Trừ (2) cho (1) vế theo vế ta được:

\(\dfrac{1}{x}\) + \(\dfrac{2}{y}\) - \(\dfrac{1}{x}\) - \(\dfrac{1}{y}\) = \(\dfrac{1}{12}\) - \(\dfrac{1}{16}\)

<=> \(\dfrac{1}{y}\) = \(\dfrac{1}{48}\) <=> y = 48

Thay y =48 vào (2) ta có: \(\dfrac{1}{x}\) + \(\dfrac{1}{48}\) = \(\dfrac{1}{16}\)

<=> \(\dfrac{1}{x}\) = \(\dfrac{1}{24}\) <=> x = 24

Vậy x =24 ; y =48

Ta có :

\(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+....+\frac{1}{\left(x+5\right)\left(x+6\right)}\)

\(=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+....+\frac{1}{x+5}-\frac{1}{x+6}\)

\(=\frac{1}{x}-\frac{1}{x+6}\)

\(=\frac{6}{x\left(x+6\right)}\)

\(CMR\) \(a^2+b^2+c^2\ge\frac{1}{3}\)

\(\frac{1}{x}\)+\(\frac{1}{y}\)=\(\frac{1}{24}\)<=>\(\frac{24y}{24xy}\)+\(\frac{24x}{24xy}\)=\(\frac{xy}{24xy}\)

<=> 24y +24x=xy<=> (24y-xy) -(576-24x)+576=0

<=> y(24-x) -24(24-x)=-576

<=> (24-x)(y-24)=-576=-576.1=1.(-576)=(-24).24=24.(-24)=12.(-48)=48.(-12)=....

và lần lượt cho 24-x và y-24 = các kết quả kia và chỉ lấy những giá trị là số tự nhiên

Mày nhìn cái chóa j

Ta có : 

\(\Rightarrow2\left(5x-2\right)=3\left(5-3x\right)\)

\(\Leftrightarrow10x-4=15-9x\)

\(\Leftrightarrow10x+9x=15+4\)

=> 19x = 19

=> x = 1

Ta có : 

\(\Leftrightarrow\frac{10x+3}{12}=\frac{9}{9}+\frac{6+8x}{9}\)

\(\Leftrightarrow\frac{10x+3}{12}=\frac{15+8x}{9}\)

=> (10x + 3)9 = (15 + 8x).12

=> 90x + 27 = 180 + 96x

=> 90x - 96x = 180 - 27

=> -6x = 153

=> -x = 25,5

=> x = -25,5

a) \(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}=\sqrt{3}-2\)

b) \(\left(\frac{1}{x-4}-\frac{1}{x+4\sqrt{x}+4}\right).\frac{x+2\sqrt{x}}{\sqrt{x}}=\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{1}{\left(\sqrt{x}+2\right)^2}\right).\left(\sqrt{x}+2\right)\)

\(=\frac{\sqrt{x}+2-\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}.\left(\sqrt{x}+2\right)=\frac{4}{x-4}\)

a, \(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}=\sqrt{3}-\sqrt{4}\)

b, Với x > 0 ; x \(\ne\)4

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

\(=\left(\frac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{1}{\left(\sqrt{x}+2\right)^2}\right)\left(\sqrt{x}+2\right)\)

\(=\frac{\sqrt{x}+2}{\left(\sqrt{x}\pm2\right)}-\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2}=\frac{1}{\sqrt{x}-2}-\frac{1}{\sqrt{x}+2}\)

\(=\frac{\sqrt{x}+2-\sqrt{x}+4}{\left(\sqrt{x}\pm2\right)}=\frac{6}{\left(\sqrt{x}\pm2\right)}\)

1.Với \(x-1\ge0\Rightarrow x\ge1\)

\(\Rightarrow x^2-3x+2+x-1=0\Rightarrow x^2-2x+1=0\)

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

Với \(x-1< 0\Rightarrow x< 1\)

\(\Leftrightarrow x^2-3x+2-x+1=0\Leftrightarrow x^2-4x+3=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}\left(l\right)}\)

Vậy x=1

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

ĐK \(x\ne0\)và\(x\ne2\)

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

\(\Rightarrow x^2+x=0\Rightarrow x\left(x+1\right)=0\Rightarrow\orbr{\begin{cases}x=0\left(l\right)\\x=-1\left(tm\right)\end{cases}}\)

Vậy x=-1

Tam giác AMB đồng dạng với tam giác BMN ( Tự chứng minh )

Suy ra \(\frac{AM}{BM}=\frac{AD}{BN}\Rightarrow AM.BN=AD.BM\)

b) Ta chứng minh tam giác ADM bằng tam giác CDK

Rồi suy ra tam giác DMK cân

Mà DM vuông góc với DK

Nên tam giác DMK vuông cân

1. \(x^2-3x+2\) + / x - 1 / = 0 ( 1)

+) Với : x ≥ 1 , ta có :

( 1) ⇔ x² - 3x + 2 + x - 1 = 0

⇔ x² - 2x + 1 = 0

⇔ ( x - 1)² = 0

⇔ x = 1 ( TM ĐK )

+) Với : x < 1 , ta có :

( 1) ⇔ x² - 3x + 2 + 1 - x = 0

⇔ x² - 4x + 3 = 0

⇔ x² - x - 3x + 3 = 0

⇔ x( x - 1) - 3( x - 1) = 0

⇔ ( x - 1)( x - 3) = 0

⇔ x = 1 ( KTM ) hoặc : x = 3 ( KTM )

KL.......

3. \(\dfrac{x+2}{x-2}-\dfrac{1}{x}-\dfrac{2}{x\left(x-2\right)}=0\) ( x # 2 ; x # 0)

⇔ \(\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}-\dfrac{2}{x\left(x-2\right)}=0\)

⇔ x² + 2x + 2 - x - 2 = 0

⇔ x² + x = 0

⇔ x( x + 1) = 0

⇔ x = 0 ( KTM) hoặc : x = -1 ( TM )

KL....