\(\frac{2}{5}.\frac{1}{x}+\frac{1}{x}.2+\frac{2}{5}=0,5\)

\(\Rightarrow\frac{2}{5x}+\frac{2}{x}+\frac{2}{5}=\frac{1}{2}\)

\(\Rightarrow2.\left(\frac{1}{5x}+\frac{1}{x}+\frac{1}{5}\right)=\frac{1}{2}\)

\(\Rightarrow\frac{1}{5x}+\frac{5}{5x}+\frac{x}{5x}=\frac{1}{2}:2=\frac{1}{4}\)

\(\Rightarrow\frac{1+5+x}{5x}=\frac{1}{4}\)

\(\Rightarrow4.\left(1+5+x\right)=5x\)

\(\Rightarrow4+20+4x=5x\)

\(\Rightarrow24+4x=5x\)

\(\Rightarrow5x-4x=24\)

\(\Rightarrow x=24\)

Trả lời

2/5.1/x+1/x.2+2/5=0,5

1/x.(2/5+2)+2/5   =1/2

1/x.12/5+2/5        =1/2

1/x.12/5               =1/2-2/5

1/x.12/5               =1/10

1/x                       =1/10:12/5

1/x                       =1/24.

Hình như mk làm sai !

\(\frac{2}{5}.\frac{1}{x}+\frac{1}{x}.2+\frac{2}{5}=0,5\)

\(\Rightarrow\frac{1}{x}\left(\frac{2}{5}+\frac{4}{2}\right)=\frac{1}{2}-\frac{2}{5}\)

\(\Rightarrow\frac{1}{x}.\frac{12}{5}=\frac{1}{10}\)

\(\Rightarrow\frac{1}{x}=\frac{1}{24}\)

\(\Rightarrow x=24\)

~Study well~

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

b)\(\left(x-8\right)\left(x-2\right)=0\Leftrightarrow\orbr{\begin{cases}x-8=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=2\end{cases}}\)

c) \(\left(x+1\right)+\left(x+2\right)+...+\left(x+10\right)=9x+200\)

\(\Leftrightarrow\left(x+x+...+x\right)+\left(1+2+...+10\right)=9x+200\) (10 số hạng x)

\(\Leftrightarrow10x+55=9x+200\Leftrightarrow x+55=200\)

\(\Leftrightarrow x=145\)

các bạn ơi giúp mik ý đầu tiên đi

các bạn ơi giúp mik với mik đang cần gấp ai làm nhanh mik sẽ k giải chi tiết giúp mik nhé cảm ơn

(125^3 .7^5 -17^5 ) : 2001^2

(x-8) .(x-2) =0

(x+1) +(x+2)+...........+(x+10)=9.x+200 

3/5 - 1/3 x (2,48 + 0,52) x y : 60 : 5 = 1/5

       1/3 x (2,48+0,52) x y : 60 : 5 = 2/5

    1/3 x 3 x y : 60 : 5                     = 2/5

                      y: 60 : 5                    =2/5

                      y: 60                          =2/5 x 5

                      y : 60                         =2 

                      y                                =  2 x 60

                      y                                  =120

Ta có: \(\dfrac{3}{5}-\dfrac{1}{3}\left(2.48+0.52\right)\cdot y:60:5=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{3}{5}-\dfrac{1}{3}\cdot3\cdot y\cdot\dfrac{1}{60}\cdot\dfrac{1}{5}=\dfrac{1}{5}\)

\(\Leftrightarrow y\cdot\dfrac{1}{300}=\dfrac{2}{5}\)

hay y=120

\(x\times\frac{4}{7}+\frac{1}{3}\times x=\frac{2}{3}\)

\(x\times\left(\frac{4}{7}+\frac{1}{3}\right)=\frac{2}{3}\)

\(x\times\frac{19}{21}=\frac{2}{3}\)

\(x=\frac{2}{3}:\frac{19}{21}\)

\(x=\frac{2}{3}\times\frac{21}{19}=\frac{14}{19}\)

cảm ơn bé sói nha

C1

a) -7x(3x-2)=-21x^2+14x

b) 87^2+26.87+13^2=87^2+2.13.87+13^2=(87+13)^2=100^2

C2

a) (x-5)(x+5)

b)3x(x+5)-2(x+5)=(3x-2)(x+5)=0

\(\Rightarrow\left[\begin{array}{nghiempt}3x-2=0\\x+5=0\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=\frac{2}{3}\\x=-5\end{array}\right.\)

Vậy S={-5;2/3}

C3:

a)3x^3-2x^2+2=(x+1)(3x^2-5x-5)-3

b) Để A chia hết cho B=> x+1\(\inƯ\left(-3\right)\)

\(\Rightarrow\begin{cases}x+1=3\\x+1=-3\\x+1=1\\x+1=-1\end{cases}\)\(\Rightarrow\begin{cases}x=2\\x=-4\\x=0\\x=-2\end{cases}\)

a) 3x3-2x2+2 chia x+1= 3x2-5x+5 dư -3 b) -3 chia hết x+1 vậy chon x =2

1)

a) \(-7x\left(3x-2\right)\)

\(=-21x^2+14x\)

b) \(87^2+26.87+13^2\)

\(=87^2+2.87.13+13^2\)

\(=\left(87+13\right)^2\)

\(=100^2\)

\(=10000\)

2)

a) \(x^2-25\)

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

\(=\left(x-5\right)\left(x+5\right)\)

b) \(3x\left(x+5\right)-2x-10=0\)

\(\Leftrightarrow3x\left(x+5\right)-\left(2x-10\right)=0\)

\(\Leftrightarrow3x\left(x+5\right)-2\left(x-5\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{2}{3}\end{matrix}\right.\)

Vậy..........

3)

a) \(A:B=\left(3x^3-2x^2+2\right):\left(x+1\right)\)

Vậy \(\left(3x^3-2x^2+2\right):\left(x+1\right)=\left(3x^2-5x-5\right)+7\)

b)

Để \(A⋮B\Rightarrow7⋮\left(x+1\right)\)

\(\Rightarrow\left(x+1\right)\in U\left(7\right)=\left\{-1;1-7;7\right\}\)

Vì x là số nguyên nên x=0 ; x=6 thì \(A⋮B\)

Đ/S\(=\frac{703}{253}\)

chúc bạn hk tốt!!!

Chịu thôi!Tớ biết cách làm bài nhưng đến phép cuối thì ko so sánh được!Cậu xem lại đầu bài nha!

Giải ra luôn nha

\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+.......+\dfrac{1}{x\cdot\left(x+1\right)}=\dfrac{122}{123}\)

\(\Leftrightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+......+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{122}{123}\)

\(\Leftrightarrow1-\dfrac{1}{x+1}=\dfrac{122}{123}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{123}\)

\(\Leftrightarrow x=122\)