Chắc câu b sai?

\(\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{6}=\)

\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}=\)

\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=\)

\(\frac{1}{1}-\frac{1}{6}=\frac{5}{6}\)

k nha gõ mỏi tay lắm

149/240

nguyễn trần đức kông bn phải giải rõ ràng ra chứ, 

149/240

trả lời ở trên

[(x-1):1+1].(x+1):2=1450

x.(x+1):2=1450

x.(x+1)=2900

mà x và x+1 là hai stn liên tiếp mà 2900 ko là tích của hai stn liên tiếp => ko có gtrij x tm

( x + 1 ) . [ (x - 1 ) :1 + 1 ) ] : 2 = 1450

( x + 1 ) . x : 2 = 1450

( x + 1 ) . x  = 1450 . 2

( x + 1 ) . x = 2900

Mà x  và x + 1 là hai số tự nhiên liên tiếp  2900 ko là tích của hai số tụ nhiên liên tiếp . => x ko có giá trị

ai giúp mk bài này vs có ai giải nữa ko zậy

1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x + 1) = 99/100
1- 1/2 +1/2-1/3+1/3-1/4+...+ 1/x - 1/ x+ 1 = 99/100
1 - 1/ x+1 = 99/ 100
=> (100 - 1)/ x+1 = 99 / 100
=> x+1 = 100 => x=99

\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{99}{100}\)

\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{99}{100}\)

\(\Rightarrow1-\frac{1}{x+1}=\frac{99}{100}\)

\(\Rightarrow\frac{1}{x+1}=1-\frac{99}{100}=\frac{1}{100}\)

\(\Rightarrow x+1=100\)

\(\Rightarrow x=99\)

mình sửa đề mới làm đc cái chỗ 1/2 phải là 1/1.2 ( đúng ko . xem lại )

A = 1/1.2 + 1/2.3 +....+ 1/x.(x+1)=99/100

A=1/1 - 1/2 + 1/2 - 1/3 +...+ 1/x - 1/x+1 =99/100

A = 1 - 1/x+1 = 99/100

A=x+1 - 1/x+1 = 99/100

A=x/x+1 = 99/100

=> x=99

\(\left(x+2\right)\left(x^2+2x-9\right)\)

\(=x^3+2x^2-9x+2x^2+4x-18\)

\(=x^3+4x^2-5x-18\)

\(\left(x^{2y}-6\right)\left(x^2-5\right)\)

\(=x^{4y}-5x^{2y}-6x^2+30\)

\(\left(x+y\right)\left(xy-4+y\right)\)

\(=x^2y-4x+xy+xy^2-4y+y^2\)

câu còn lại tương tự  nha

1) Ta có: \(4x^2-1=\left(2x+1\right).\left(3x-5\right)\)

\(\Leftrightarrow\left(2x+1\right).\left(2x-1\right)-\left(2x+1\right).\left(3x-5\right)=0\)

\(\Leftrightarrow\left(2x+1\right).\left[\left(2x-1\right)-\left(3x-5\right)\right]=0\)

\(\Leftrightarrow\left(2x+1\right).\left(2x-1-3x+5\right)=0\)

\(\Leftrightarrow\left(2x+1\right).\left(4-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-1\\-x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{2}\left(TM\right)\\x=4\left(TM\right)\end{matrix}\right.\)

Vậy \(x=-\frac{1}{2}\) hoặc \(x=4\)

2) Ta có: \(\left(x+1\right)^2=4.\left(x^2-2x+1\right)\)

\(\Leftrightarrow\left(x+1\right)^2-\left[2.\left(x-1\right)\right]^2=0\)

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

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

\(\Leftrightarrow\left(3x-1\right).\left(3-x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\-x=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(TM\right)\\x=3\left(TM\right)\end{matrix}\right.\)

Vậy \(x=\frac{1}{3}\) hoặc \(x=3\)

3) Ta có: \(2x^3+5x^2-3x=0\)

\(\Leftrightarrow x.\left(2x^2+5x-3\right)=0\)

\(\Leftrightarrow x.\left(2x^2-x+6x-3\right)=0\)

\(\Leftrightarrow x.\left[x.\left(2x-1\right)+3.\left(2x-1\right)\right]=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=-3\left(TM\right)\\x=-\frac{1}{2}\left(TM\right)\end{matrix}\right.\)

Vậy \(x=0\) hoặc \(x=-3\) hoặc \(x=-\frac{1}{2}\)

4) Ta có: \(2x=3x-2\)

\(\Leftrightarrow2x-3x=-2\)

\(\Leftrightarrow-x=-2\)

\(\Leftrightarrow x=2\left(TM\right)\)

Vậy \(x=2\)

5) Ta có: \(x+15=3x-1\)

\(\Leftrightarrow x-3x=-1-15\)

\(\Leftrightarrow-2x=-16\)

\(\Leftrightarrow x=8\left(TM\right)\)

Vậy \(x=8\)

6) Ta có: \(2-x=0,5x-4\)

\(\Leftrightarrow-x-0,5x=-4-2\)

\(\Leftrightarrow-1,5x=-6\)

\(\Leftrightarrow x=4\left(TM\right)\)

Vậy \(x=4\)

1) 4x²-1=(2x+1)(3x-5)

<=> (2x-1)(2x+1)-(2x+1)(3x-5)=0

<=> (2x+1)(2x-1-3x+5)=0

<=> (2x+1)(4-x)=0

<=>\([^{2x+1=0}_{4-x=0}< =>[^{2x=-1}_{x=4}< =>[^{x=\frac{-1}{2}}_{x=4}\)

2) (x+1)²= 4(x²-2x+1)

<=> x²+2x+1-4(x²-2x+1)=0

<=> x²+2x+1-4x²+8x-4=0

<=> -3x²+10x-3=0

<=> -3x²+x+9x-3=0

<=> -x(3x-1)+3(3x-1)=0

<=> (3x-1)(3-x)=0

<=> \([^{3x-1=0}_{3-x=0}< =>[^{3x=1}_{x=3}< =>[^{x=\frac{1}{3}}_{x=3}\)

3) 2x³+5x²-3x=0

<=> 2x(x²+\(\frac{5}{2}x-\frac{3}{2})=0\)

<=> 2x\(\left[x^2+2.\frac{5}{4}x+\frac{25}{16}-\left(\frac{25}{16}+\frac{3}{2}\right)\right]=0\)

<=> 2x\(\left[\left(x+\frac{5}{4}\right)^2-\frac{49}{16}\right]=0\)

<=> 2x\(\left(x+\frac{5}{4}-\frac{7}{4}\right)\left(x+\frac{5}{4}+\frac{7}{4}\right)=0\)

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

<=>\(\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\\x=-3\end{matrix}\right.\)

4) 2x=3x-2

<=> 2x-3x=-2

<=> -x=-2

<=> x=2

5) x+15=3x-1

<=> x-3x=1-15

<=> -2x=-14

<=> x=-14:-2

<=> x=7

6) 2-x=0,5x-4

<=> -x-0,5x=-4-2

<=> -1,5x=-6

<=> x= -6: -1,5

<=> x=4

học tốt nghen

1, \(4x^2-1=\left(2x+1\right)\left(3x-5\right)\)

\(\Leftrightarrow4x^2-1=6x^2-10x+3x-5\)

\(\Leftrightarrow4x^2-1=6x^2-7x-5\)

\(\Leftrightarrow6x^2-7x-5-4x^2+1=0\)

\(\Leftrightarrow2x^2-7x-4=0\)

\(\Leftrightarrow\left(2x^2-8x\right)-\left(x-4\right)=0\)

\(\Leftrightarrow2x\left(x-4\right)-\left(x-4\right)=0\)

\(\Leftrightarrow\left(x-4\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-4=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{1}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{4;\frac{1}{2}\right\}\)

2, \(\left(x+1\right)^2=4\left(x^2-2x+1\right)\)

\(\Leftrightarrow x^2+2x+1=4x^2-8x+4\\ \Leftrightarrow4x^2-8x+4-x^2-2x-1=0\\ \Leftrightarrow3x^2-10x+3=0\\ \Leftrightarrow\left(3x^2-x\right)-\left(9x-3\right)=0\\ \Leftrightarrow x\left(3x-1\right)-3\left(3x-1\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=3\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{3;\frac{1}{3}\right\}\)

3,

\(2x^3+5x^2-3x=0\\ \Leftrightarrow x\left(2x^2+5x-3\right)=0\\ \Leftrightarrow x\left[\left(2x^2+x\right)-\left(6x+3\right)\right]=0\\ \Leftrightarrow x\left[x\left(2x+1\right)-3\left(2x+1\right)\right]=0\\ \Leftrightarrow x\left(2x+1\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}x=0\\2x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\frac{1}{2}\\x=3\end{matrix}\right.\)

Vậy phương trình có tập nghiệm là \(S=\left\{0;-\frac{1}{2};3\right\}\)

a) (-x-1/2).3/4+1=1/3

=>(-x-1/2).3/4=-2/3

=>-x-1/2=-2/3:3/4

=>-x-1/2=-8/9

=>-x=(-8/9)+1/2

=>-x=-7/18=>x=7/18

bn nào giúp mk tick cho

yêu các bn nhìu