\(\frac{1}{6}+\frac{1}{66}+\frac{1}{176}+...+\frac{1}{5396}=\frac{1}{1.6}+\frac{1}{6.11}+\frac{1}{11.16}+...+\frac{1}{71.76}\)

\(=\frac{1}{5}\left(\frac{5}{1.6}+\frac{5}{6.11}+\frac{5}{11.16}+...+\frac{5}{71.76}\right)\)

\(=\frac{1}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{71}-\frac{1}{76}\right)\)

\(=\frac{1}{5}\left(1-\frac{1}{76}\right)=\frac{1}{5}.\frac{75}{76}=\frac{25}{76}\)

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

\(\frac{4}{x-5}-\frac{1}{x+5}+\frac{13x-x^2}{25-x^2}=\frac{4}{x-5}-\frac{1}{x+5}+\frac{x^2-13x}{x^2-25}\)

\(=\frac{4\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}-\frac{x-5}{\left(x-5\right)\left(x+5\right)}+\frac{x^2-13x}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{4x+20-x+5+x^2-13x}{\left(x-5\right)\left(x+5\right)}\)

\(=\frac{x^2-10x+25}{\left(x-5\right)\left(x+5\right)}=\frac{\left(x-5\right)^2}{\left(x-5\right)\left(x+5\right)}=\frac{x-5}{x+5}\)

ui toán 6 sao ghi toán 1 chi

a) x(x - 3) + 5x = x² - 8

=> x² - 3x + 5x - x² + 8 = 0

=> 2x + 8 = 0

=> 2x = -8

=> x = -4

b) 3(x + 4) - x² - 4x = 0

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

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

=> \(\orbr{\begin{cases}3-x=0\\x+4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=-4\end{cases}}\)

Vậy \(x\in\left\{3;-4\right\}\)là giá trị cần tìm

c) 7x³ + 12x² - 4x = 0

=> x(7x² + 12x - 4) = 0

=> x(7x² + 14x - 2x - 4) = 0

=> x[7x(x + 2) - 2(x + 2)] = 0

=> x(x + 2)(7x - 2) = 0

=> x = 0 hoặc x + 2 = 0 hoặc 7x - 2 = 0

=> x = 0 hoặc x = -2 hoặc x = 2/7

Vậy \(x\in\left\{0;-2;\frac{2}{7}\right\}\)là giá trị cần tìm

x( x - 3 ) + 5x = x² - 8

⇔ x² - 3x + 5x - x² + 8 = 0

⇔ 2x + 8 = 0

⇔ 2x = -8

⇔ x = -4

3( x + 4 ) - x² - 4x = 0

⇔ 3( x + 4 ) - x( x + 4 ) = 0

⇔ ( x + 4 )( 3 - x ) = 0

⇔ x = -4 hoặc x = 3

7x³ + 12x² - 4x = 0

⇔ x( 7x² + 12x - 4 ) = 0

⇔ x( 7x² + 14x² - 2x - 4 ) = 0

⇔ x[ 7x( x + 2 ) - 2( x + 2 ) ] = 0

⇔ x( x + 2 )( 7x - 2 ) = 0

⇔ x = 0 hoặc x = -2 hoặc x=  2/7

a, \(5x^2y+10xy=5xy\left(x+2\right)\)

b, \(x^2-2xy+y^2-25=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)

c, \(x^3-8+2x\left(x-2\right)=\left(x-2\right)\left(x^2+2x+4\right)+2x\left(x-2\right)\)

\(=\left(x-2\right)\left[\left(x^2+2x+4\right)+2x\right]=\left(x-2\right)\left(x+2\right)^2\)

d, \(x^4+x^2y^2+y^4\):< 

a, 5xy(  x + 2)

b, ( x - y -5 )(x-y+5)

c,( x-2)( x+ 2)²

mong các bạn giải theo cách lớp 8

\(ĐK:x\ge1,x\inℝ\)

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

\(\Leftrightarrow x^4-2x^2+1=8x+4\Leftrightarrow x^4=2x^2+8x+3\)

\(\Leftrightarrow x^4+2x^2+1=4x^2+8x+4\Leftrightarrow\left(x^2+1\right)^2=\left(2x+2\right)^2\)

\(\Leftrightarrow x^2+1=2x+2\)hoặc \(x^2+1=-2x-2\)

Th1: \(x^2+1=2x+2\Leftrightarrow x^2-2x-1=0\Leftrightarrow\orbr{\begin{cases}x=1+\sqrt{2}\left(tmđk\right)\\x=1-\sqrt{2}\left(L\right)\end{cases}}\)

Th2: \(x^2+1=-2x-2\Leftrightarrow x^2+2x+3=0\Leftrightarrow\left(x+1\right)^2+2=0\)(Loại vì phương trình vô nghiệm với mọi x thực)

Vây phương trình có một nghiệm duy nhất là \(1+\sqrt{2}\)

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