\(\dfrac{7}{15}+\left(\dfrac{-4}{11}\right)+\dfrac{8}{15}-2,5+\dfrac{7}{15}\)

\(=\dfrac{7}{15}+\left(\dfrac{-4}{11}\right)+\dfrac{8}{15}+\left(\dfrac{-5}{2}\right)+\dfrac{7}{15}\)

\(=\left(\dfrac{7}{15}+\dfrac{8}{15}+\dfrac{7}{15}\right)+\left[\left(\dfrac{-4}{11}\right)+\left(\dfrac{-5}{2}\right)\right]\)

\(=\dfrac{22}{15}+\left(\dfrac{-63}{22}\right)\)

\(=\dfrac{-461}{330}\)

\(=\dfrac{7}{15}+\dfrac{8}{15}+\dfrac{-4}{11}-\dfrac{5}{2}+\dfrac{7}{15}=\dfrac{22}{15}-\dfrac{63}{22}=-\dfrac{461}{330}\)

=7/15+8/15+(−4)/11−5/2+7/15=(22/15−63/22)=(−461)/330

5(x-2)-125=15

=>\(5\left(x-2\right)=15+125=140\)

=>x-2=28

=>x=28+2=30

Câu trả lời:

5. (𝑥 − 2) − 125 = 15 

5.  ( x - 2)          = 125 - 15

5.  ( x - 2)          = 110

      x - 2            = 110 : 5

       x - 2           =  22

       x                 =  22 - 2

       x                 =  20

mik có hc lazi nè

link của của mik  https://lazi.vn/user/bach.bach17

có j thì kb nha

mik lớp 6

=>8(x+1/x)^2+4[(x+1/x)^2-2]^2-4[(x+1/x)^2-2](x+1/x)^2=(x+4)^2

Đặt x+1/x=a(a>=2)

=>8a^2+4[a^2-2]^2-4[a^2-2]*a^2=(x+4)^2

=>8a^2+4a^4-16a^2+16-4a^4+8a^2=(x+4)^2

=>(x+4)^2=16

=>x+4=4 hoặc x+4=-4

=>x=-8;x=0

Điều kiện: \(x\ne0\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)\left[\left(x^2+\dfrac{1}{x^2}\right)-\left(x+\dfrac{1}{x}\right)^2\right]=\left(x+4\right)^2\)

\(\Leftrightarrow8\left(x+\dfrac{1}{x}\right)^2-8\left(x^2+\dfrac{1}{x^2}\right)=\left(x+4\right)^2\\ \Leftrightarrow\left(x+4\right)^2=16\\ \Rightarrow\left\{{}\begin{matrix}x=0\\x=-8\end{matrix}\right.\)

Vì \(x\ne0\) nên \(S=\left\{-8\right\}\)

a: =(8/27+19/27)+(4/15+11/15)=1+1=2

b: =(12/13+8/13+6/13)+(2/7+5/7)=2+1=3

A. x = 2

B. \(\dfrac{3}{8}=\dfrac{6}{x}\)\(\Leftrightarrow x=\dfrac{6.8}{3}=16\)

C. x = 3

D. \(x=\dfrac{4.6}{8}=3\)

E. \(x=\dfrac{7}{3}\)

G.\(\dfrac{14}{13}=\dfrac{28}{10-x}\)

<=>\(14\left(10-x\right)=364\)

<=> 10 - x = 26 

<=> x = -16 

H. \(3\left(x+2\right)=4\left(x-5\right)\)

<=> 3x + 6  = 4x - 20 

<=> -x = -26

<=> x = 26

K. \(\dfrac{x}{2}=\dfrac{8}{x}\)

<=> \(x^2=16\)

<=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

M. \(\left(x-2\right)^2=100\)

<=> \(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)

a=2

b=16

c=3

d=3

mik chỉ biết thế này thôi(ko chắc đúng=3)

A, \(\dfrac{x}{5}=\dfrac{2}{5}\Rightarrow x=2\)

B,\(\dfrac{3}{8}=\dfrac{6}{x};\dfrac{3}{8}=\dfrac{3\cdot2}{8\cdot2}=\dfrac{6}{16}\Rightarrow x=16\)

C,\(\dfrac{1}{9}=\dfrac{x}{27};\dfrac{1}{9}=\dfrac{1\cdot3}{9\cdot3}=\dfrac{3}{27}\Rightarrow x=3\)

D,\(\dfrac{4}{x}=\dfrac{8}{6};\dfrac{4}{x}=\dfrac{4\cdot2}{x\cdot2}=\dfrac{8}{2x}\)

   \(\Rightarrow2x=6;6=x:2;x=3\)

các bạn ơi giúp mình với ạ

bang 67 

a).\(\dfrac{x}{15}\)=\(\dfrac{4}{3}\)\(\Leftrightarrow\)x.3=15.4\(\Leftrightarrow\)3x=60\(\Leftrightarrow\)x=20

b)x:\(\dfrac{4}{3}\)=\(\dfrac{5}{6}\)\(\Leftrightarrow\)\(\dfrac{3x}{4}=\dfrac{5}{6}\)\(\Leftrightarrow\)3x.6=4.5\(\Leftrightarrow\)18x=20\(\Leftrightarrow\)\(\dfrac{10}{9}\)

c)\(\dfrac{7}{2}\)-x=\(\dfrac{5}{6}\)\(\Leftrightarrow\)\(\dfrac{7}{2}-\dfrac{2x}{2}\)=\(\dfrac{5}{6}\)\(\Leftrightarrow\)\(\dfrac{7-2x}{2}\)=\(\dfrac{5}{6}\)\(\Leftrightarrow\)(7-2x).6=2.5\(\Leftrightarrow\)42-12x=10

\(\Leftrightarrow\)-12x=-32\(\Leftrightarrow\)x=\(\dfrac{8}{3}\)

Ta có: \(\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{2011^2}\)

\(=\left(\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}\right)+\left(\frac{1}{8^2}+\frac{1}{9^2}+\frac{1}{10^2}+...+\frac{1}{2011^2}\right)\)

\(>\frac{1}{3^2}+\left(\frac{1}{7^2}+\frac{1}{7^2}+\frac{1}{7^2}+...+\frac{1}{7^2}\right)\)(2007 phân số \(\frac{1}{7^2}\))

\(=\frac{1}{3^2}+\left(\frac{1.2007}{7^2}\right)=\frac{1}{3^2}+\frac{2007}{7^2}>\frac{125}{503}^{\left(đpcm\right)}\)

Đặt S= 1/4^2+1/5^2=1/6^2+...+1/2011^2

Ta có: 1/3.4>1/4^2

1/4.5>1/5^2

.........

1/2010.2011>1/2011^2

Suy ra: S>1/3.4+1/4.5+1/5.6+...+1/2010.2011

S>1/3 -1/4+1/4-1/5+...+1/2010-1/2011

S>1/3-1/2011

S>2008/6033>125/503

từ đó suy ra S.125/503

k cho mình nha