Sửa đề \(\frac{11}{13}\)chứ không phải \(\frac{11}{3}\)

\(\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}-x-\frac{1}{9}=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)

+) Đặt \(A=\frac{2,75-2,2+\frac{11}{7}+\frac{11}{13}}{0,75-0,6+\frac{3}{7}+\frac{3}{13}}\)

\(A=\frac{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}\)

\(A=\frac{11\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{3\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)

\(A=\frac{11}{3}\)(1)

+) Đặt \(B=\frac{2}{3}+\frac{2}{15}+\frac{2}{35}+\frac{2}{63}\)

\(B=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}\)

\(B=\frac{2}{2}\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}\right)\)

\(B=\frac{2}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{7}-\frac{1}{9}\right)\)

\(B=\frac{2}{2}\left(1-\frac{1}{9}\right)=1\cdot\frac{8}{9}=\frac{8}{9}\)(2)

Từ (1) và (2) => \(A-x-\frac{1}{9}=B\)

=> \(\frac{11}{3}-x-\frac{1}{9}=\frac{8}{9}\)

=> \(\frac{11}{3}-x=1\)

=> \(x=\frac{11}{3}-1=\frac{8}{3}\)

Vậy x = 8/3

\(\Leftrightarrow\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\Rightarrow x\in\left(-2;-1;0;1;2\right)\)

\(\Leftrightarrow\frac{-1}{24}\le\frac{x}{24}\le\frac{5}{24}\Rightarrow x\in\left(-1;0;1;2;3;4;5\right)\)

2 câu sau tự làm nha

\(-\frac{5}{17}+\frac{3}{17}\le\frac{x}{17}\le\frac{13}{17}+-\frac{11}{17}\)

\(\frac{-2}{17}\le\frac{x}{17}\le\frac{2}{17}\)

=> \(x\in\left\{-2;-1;0;1;2\right\}\)

Còn mình nè

Lý 7đ ; Sinh 9đ ; Sử 9đ ; Địa 8,5đ ; Gdcd 9đ ; Công nghệ 8đ

Chả biết có đc HSG không nữa

a) B = \(\frac{1}{2}.\frac{4}{3}.20.\frac{7}{35}.0,75\)

B = \(\frac{1.1.2.1.1}{1.1.1.1.1}\)

B= 2

b) \(\frac{1}{3}x=16\frac{1}{4}-13\frac{1}{4}\)

= > \(\frac{1}{3}x=3\)

= > x = 9

c)\(\frac{2}{3}x=\frac{-1}{4}-\frac{1}{3}\)

=> \(\frac{2}{3}x=-\frac{7}{12}\)

=> x = \(\frac{-7}{12}:\frac{2}{3}\)

=> x = \(\frac{-7}{8}\)

sao dễ vậy

Dễ mà bn.

bài 1 cậu đổi ra xong nhân từng cặp 1laij với nhau

bài 2 cậu đổi ra sau đó cậu lấy \(\frac{65}{4}-\frac{53}{4}=3\) sau đó lấy 3:1/3

b, cậu lấy -1/4 -1/3 =-7/12 rồi lấy -7/12:2/3 =-7/8

bài 1

A= \(\frac{1}{2}.\frac{4}{3}.20.\frac{7}{35}.0,75=2\)

bài 2

a,\(\Leftrightarrow\frac{1}{3}x+\frac{53}{4}=\frac{65}{4}\)

\(\Leftrightarrow\frac{1}{3}x=3\)

\(\Leftrightarrow x=9\)

vậy....

b, \(\Leftrightarrow\frac{2}{3}x=\frac{-7}{12}\)

\(\Leftrightarrow x=-\frac{7}{8}\)

vậy.......

chuvh7uuyj

\(\frac{2}{3}+\frac{8}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)

\(\frac{94}{105}< \frac{x}{105}< \frac{92}{105}\)

\(\Rightarrow94< x< 92\)

mà x là số tựu nhiên => \(x\in\varnothing\)

\(\frac{x-7}{3}=\frac{-27}{7-x}\)

\(\Leftrightarrow\left(x-7\right)\left(7-x\right)=\left(-27\right).3\)

\(\Leftrightarrow7x-x^2-49+7x=-81\)

\(\Leftrightarrow-x^2+14x-49=-81\)

\(\Leftrightarrow-x^2+14x+32=0\)

\(\Leftrightarrow x^2-14x-32=0\)

\(\Leftrightarrow x^2+2x-16x-32=0\)

\(\Leftrightarrow x\left(x+2\right)-16\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x-16=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=16\end{cases}}}\)

x= 16 nhé

Dựa vào đề bài

\(\Rightarrow\left(-35\right).2=7.\left(x+12\right)\)

\(\Leftrightarrow\left(-70\right)=7x+84\)

\(\Leftrightarrow7x+84=\left(-70\right)\)

\(\Rightarrow7x=\left(-70\right)-84=\left(-154\right)\)

\(\Rightarrow x=\left(-154\right):7=\left(-22\right)\)

KL: x = (-22)

mink nhanh đó bạn

-35.2=(x+12).7
   -70=7x+84
 -154=7x
     x=-154:7
     x=-22
Vậy x= -22