\(a,4\frac{5}{9}:\frac{\left(-5\right)}{7}+\frac{4}{9}:\frac{-5}{7}\)

\(=\frac{41}{9}.\frac{-7}{5}+\frac{4}{9}.\frac{-7}{5}\)

\(=\frac{-7}{5}.\left(\frac{41}{9}+\frac{4}{9}\right)\)

\(=-\frac{7}{9}.5\)

\(=-7\)

a)Bn Kaito Kid làm rùi!

B)Không viết lại đề

\(=\frac{11}{7}\cdot\left(-\frac{3}{5}+\frac{4}{9}-\frac{2}{5}+\frac{5}{9}\right)=\frac{11}{7}\cdot0=0\)

c)Không viết lại đề

\(A=\left(2+4+...+100\right)\left(\frac{3}{5}\cdot\frac{10}{7}-\frac{6}{7}\right):\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)\)

\(=\left(2+4+6+...+100\right)\cdot0\cdot\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{100}\right)=0\)

\(=\frac{7}{6}\cdot\left(\frac{3}{26}-\frac{3}{13}+\frac{1}{10}-\frac{8}{5}\right)=\frac{7}{6}\left(\frac{-3}{26}+\frac{-17}{10}\right)=\frac{7}{6}\cdot\frac{236}{130}=\frac{413}{195}\)

D)

Dòng trên chữ D) là phần c)

Dưới phần c) là phần d) nha!(Lỗi đánh máy)

a= (\(\frac{2}{5}\)+\(\frac{2}{9}\)+\(\frac{2}{11}\)\(\times\)\(\frac{5}{7}\)\(+\frac{7}{9}\)\(+\frac{7}{11}\)\()\)

(7x6=5+2+6x7) =

Bấm máy tính:

E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{5}:\frac{4}{5}\)

E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{4}\)

E = \(\frac{7}{3}\)

Vậy E = \(\frac{7}{3}\)

a)\(\frac{2}{3}:\frac{4}{9}+\frac{1}{3}:\frac{4}{9}\)

=\(\frac{4}{9}:\left(\frac{2}{3}+\frac{1}{3}\right)\)

=\(\frac{4}{9}:1=\frac{4}{9}\)

b

a 9/4

b 49/40

c 36/35

d 144/441

\(E=\frac{0,8:\left(\frac{4}{5}.1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right).2\frac{2}{17}}+\left(1,2.0,5\right):\frac{4}{5}\)

\(E=\frac{\frac{4}{5}:\frac{4}{5}:1,25}{\frac{16}{25}-\frac{1}{25}}+\frac{\left(\frac{27}{25}-\frac{2}{25}\right).\frac{7}{4}}{\left(\frac{59}{9}-\frac{13}{4}\right).\frac{36}{17}}+\frac{6}{5}.\frac{1}{2}.\frac{5}{4}\)

\(E=\frac{1:\frac{5}{4}}{\frac{3}{5}}+\frac{1.\frac{7}{4}}{\frac{119}{36}.\frac{36}{17}}+\frac{3}{4}\)

\(E=\frac{4}{5}.\frac{5}{3}+\frac{\frac{7}{4}}{7}+\frac{3}{4}\)

\(E=\frac{4}{3}+\frac{7}{4}.\frac{1}{7}+\frac{3}{4}\)

\(E=\frac{4}{3}+\frac{1}{4}+\frac{3}{4}\)

\(E=\frac{4}{3}+1=\frac{7}{3}\)

1) Vì theo đề bài \(\frac{x-2}{x-6}>0\Rightarrow x\ne0\)

Gọi phân số là \(\frac{a}{b}\)với \(a>b\) (vì tử số lớn hơn mẫu số thì phân số sẽ lớn hơn 1)

 \(\Rightarrow x\ge6\)

2) Ta có: \(\frac{3x+9}{x-4}\) có giá trị nguyên . Với 3x + 9 > x - 4

Nếu x = 1 thì \(\frac{3x+9}{x-4}=\frac{31+9}{1-4}=\frac{40}{-31,3333}\) (loại)

Nếu x = 2 thì \(\frac{3x+9}{x-4}=\frac{32+9}{2-4}=\frac{41}{-2}=-20,5\) (loại)

Nếu x = 3 thì \(\frac{3x+9}{x-4}=\frac{33+9}{3-4}=\frac{42}{-1}=-42\)(chọn)

Nếu x = 4 thì \(\frac{3x+9}{x-4}=\frac{34+9}{4-4}=\frac{43}{0}\)(chọn)

Nếu x = 5 thì \(\frac{3x+9}{x-4}=\frac{35+9}{5-4}=\frac{44}{1}=44\)chọn

..và còn nhiều giá trị khác nữa...

Suy ra x = {-3 ; -4 ; -5 ; 3 ; 4 ; 5 ...}Tương tự ta có bảng sau:

Bài 3. Bí rồi, mình mới lớp 6 thôi!

bài 3: đạt B=\(\frac{1}{2}:\left(-1\frac{1}{2}\right):1\frac{1}{3}:\left(-1\frac{1}{4}\right):1\frac{1}{5}:\left(-1\frac{1}{6}\right)\):...:\(\left(-1\frac{1}{100}\right)\)

=\(\frac{1}{2}:\frac{-3}{2}:\frac{4}{3}:\frac{-5}{4}:\frac{6}{5}:\frac{-7}{6}:...:\frac{-101}{100}\)=\(\frac{1}{2}.\frac{-2}{3}.\frac{3}{4}.\frac{-4}{5}.\frac{5}{6}\frac{-6}{7}...\frac{-100}{101}\)(có 50 thừa số âm)

=\(\frac{1.2.3.4...100}{2.3.4...101}=\frac{1}{101}\)

vậy B=\(\frac{1}{101}\)

#HỌC TỐT#

   \(\frac{1}{2}\div\left(-1\frac{1}{2}\right)\div1\frac{1}{3}\div\left(-1\frac{1}{4}\right)\div1\frac{1}{5}\div...\div1\frac{1}{99}\div\left(-1\frac{1}{100}\right)\)

\(=\frac{1}{2}\div\left(-\frac{3}{2}\right)\div\frac{4}{3}\div\left(-\frac{5}{4}\right)\div\frac{6}{5}\div...\div\frac{100}{99}\div\left(-\frac{101}{100}\right)\)

\(=\frac{1}{2}.\frac{-2}{3}.\frac{3}{4}.\frac{-4}{5}.\frac{5}{6}...\frac{99}{100}.\frac{-100}{101}\)

\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}.\frac{5}{6}...\frac{99}{100}.\frac{100}{101}\)                 ( do có 50 thừa số âm )

\(=\frac{1.2.3.4.5...99.100}{2.3.4.5.6...100.101}\)

\(=\frac{1}{101}\)

\(\Rightarrow\left(\frac{x+1}{9}+1\right)+\left(\frac{x+2}{8}+1\right)=\left(\frac{x+3}{7}+1\right)+\left(\frac{x+4}{6}+1\right)\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}=\frac{x+10}{7}+\frac{x+10}{6}\)

\(\Rightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Rightarrow\left(x+10\right).\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

\(\Rightarrow x+10=0\Rightarrow x=-10\)

\(a)\frac{62}{7}\cdot x=\frac{29}{9}\div\frac{3}{56}\)

\(\Rightarrow\frac{62}{7}\cdot x=\frac{29}{9}\cdot\frac{56}{3}\)

\(\Rightarrow\frac{62}{7}\cdot x=\frac{1624}{27}\)

\(\Rightarrow x=\frac{1624}{27}\div\frac{62}{7}\)

\(\Rightarrow x=\frac{1624}{27}\cdot\frac{7}{62}\)

\(\Rightarrow x=\frac{11368}{1674}=\frac{5684}{837}\)

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