\(\frac{4}{5}\)x \(\frac{8}{3}\) x\(\frac{x}{7}\) =\(\frac{96}{105}\)\(\frac{7}{4}\)x \(\frac{3}{x}\) x\(\frac{11}{5}\) =\(\frac{231}{200}\) de bai la tim x\(\frac{3}{4}\) -\(\frac{2}{5}\) \(\frac{5...

#Toán lớp 4

1

DL

Đức Lộc

4 tháng 3 2019

1) \(\frac{4}{5}.\frac{8}{3}.\frac{x}{7}=\frac{96}{105}\)

\(\frac{32.x}{105}=\frac{96}{105}\)

\(32x=96\)

\(x=3\)

Giải các phương trình saua, 5-(x-6)=4(3-2x)b, 3-4x(25-2x)=8x2+x-300c, x-\(\frac{2x-5}{5}+\frac{x+8}{6}=7+\frac{x-1}{3}\)d, \(\frac{x+1}{15}+\frac{x+2}{7}\frac{x+4}{4}+6=0\)e,\(\frac{x-91}{37}+\frac{x-86}{42}+\frac{x-78}{50}+\frac{x-49}{79}=4\)g, \(\frac{x+14}{200}+\frac{x+27}{187}+\frac{x+105}{109}=\frac{x+200}{14}+\frac{x+187}{27}+\frac{x+109}{105}\)Các bạn giúp mk...

VP

Vũ Phương Anh

6 tháng 2 2019

#Toán lớp 8

6

GT

ღᏠᎮღ🅼🅸🅽🅷ঔ🅽🅶ụ🆈ঌ__[ᴅʀᴇᴀм тᴇᴀм]__

6 tháng 2 2019

\(a)5-\left(x-6\right)=4\left(3-2x\right)\)

\(\Leftrightarrow5-x+6=12-8x\)

\(\Leftrightarrow-x+8x=12-5-6\)

\(\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)

HB

Hoàng Băng Nhi

6 tháng 2 2019

a) 5-(x-6)=4(3-2x)

<=>5-x-6=12-8x

<=>-x+8x=2-5-6

<=>7x=1

<=>x=1/7

LT

Lê Thu Hằng

28 tháng 9 2017

Tìm x, biết:

a) \(\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}=\frac{x+5}{11}+\frac{x+5}{13}\)

b)\(\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}\)

c) (x+2) - (x+3) >0

d)\(\left(x-5\right)\left(x+\frac{7}{3}\right)\ge0\)

2

PT

Phạm Tú Uyên

28 tháng 9 2017

a) \(\dfrac{x+5}{5}+\dfrac{x+5}{7}+\dfrac{x+5}{9}=\dfrac{x+5}{11}+\dfrac{x+5}{13}\)

\(\Rightarrow\left(x+5\right)\left(\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}\right)=\left(x+5\right)\left(\dfrac{1}{11}+\dfrac{1}{13}\right)\)

\(\Rightarrow\dfrac{143}{315}\left(x+5\right)=\dfrac{24}{143}\left(x+5\right)\)

\(\Rightarrow\dfrac{143}{315}\left(x+5\right)-\dfrac{24}{143}\left(x+5\right)=0\)

\(\Rightarrow\left(x+5\right)\left(\dfrac{143}{315}-\dfrac{24}{143}\right)=0\)

\(\Rightarrow x+5=0\Rightarrow x=-5\)

b) \(\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)

\(\Rightarrow\)\(3+\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=3+\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)

\(\Rightarrow\)\(1+\dfrac{x+2}{100}+1+\dfrac{x+3}{99}+1+\dfrac{x+4}{98}=1+\dfrac{x+5}{97}+1+\dfrac{x+6}{96}+1+\dfrac{x+7}{95}\)

\(\Rightarrow\)\(\dfrac{100}{100}+\dfrac{x+2}{100}+\dfrac{99}{99}+\dfrac{x+3}{99}+\dfrac{98}{98}+\dfrac{x+4}{98}=\dfrac{97}{97}+\dfrac{x+5}{97}+\dfrac{96}{96}+\dfrac{x+6}{96}+\dfrac{95}{95}+\dfrac{x+7}{95}\)\(\Rightarrow\)\(\dfrac{x+102}{100}+\dfrac{x+102}{99}+\dfrac{x+102}{98}=\dfrac{x+102}{97}+\dfrac{x+102}{96}+\dfrac{x+102}{95}\)

\(\Rightarrow\)\(\left(x+102\right)\left(\dfrac{1}{100}+\dfrac{1}{99}+\dfrac{1}{98}\right)=\left(x+102\right)\left(\dfrac{1}{97}+\dfrac{1}{96}+\dfrac{1}{95}\right)\)

\(\Rightarrow\)\(x+102=0\)

\(\Rightarrow x=-102\)

PT

Phạm Tú Uyên

28 tháng 9 2017

c) \(\left(x+2\right)-\left(x+3\right)>0\)

\(\Rightarrow x+2-x-3>0\Rightarrow-1>0\)

\(\Rightarrow x\in\varnothing\)

d) \(\left(x-5\right)\left(x+\dfrac{7}{3}\right)\ge0\)

TH1: \(\left\{{}\begin{matrix}x-5\ge0\\x+\dfrac{7}{3}\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge5\\x\ge\dfrac{-7}{3}\end{matrix}\right.\)

\(\Rightarrow x\ge\dfrac{-7}{3}\)

TH2: \(\left\{{}\begin{matrix}x-5\le0\\x+\dfrac{7}{3}\le0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\le5\\x\le\dfrac{-7}{3}\end{matrix}\right.\)

\(\Rightarrow x\le5\)

TH3: \(\left[{}\begin{matrix}x-5=0\\x+\dfrac{7}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-7}{3}\end{matrix}\right.\)

LT

Lê Thu Hằng

28 tháng 9 2017

Tìm x, biết:

a) \(\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}=\frac{x+5}{11}+\frac{x+5}{13}\)

b)\(\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}\)

c) (x+2) - (x+3) >0

d)\(\left(x-5\right)\left(x+\frac{7}{3}\right)\ge0\)

1

NT

Nguyễn Trần Thiên Chi

28 tháng 9 2017

a) Ta có : \(\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}=\frac{x+5}{11}+\frac{x+5}{13}\)

\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\left(\frac{x+5}{11}+\frac{x+5}{13}\right)=0\)

\(\Rightarrow\frac{x+5}{5}+\frac{x+5}{7}+\frac{x+5}{9}-\frac{x+5}{11}-\frac{x+5}{13}=0\)

\(\Rightarrow\left(x+5\right)\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\right)=0\)

Do \(\frac{1}{5}+\frac{1}{7}+\frac{1}{9}-\frac{1}{11}-\frac{1}{13}\ne0\)

\(\Rightarrow x+5=0\Rightarrow x=-5\)

Vậy x = -5

b) Ta có : \(\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}\)

\(\Rightarrow\frac{x+2}{100}+\frac{x+3}{99}+\frac{x+4}{98}+3=\frac{x+5}{97}+\frac{x+6}{96}+\frac{x+7}{95}+3\)

\(\Rightarrow\frac{x+2}{100}+1+\frac{x+3}{99}+1+\frac{x+4}{98}+1=\frac{x+5}{97}+1+\frac{x+6}{96}+1+\frac{x+7}{95}+1\)

\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}=\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\)

\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\left(\frac{x+102}{97}+\frac{x+102}{96}+\frac{x+102}{95}\right)=0\)

\(\Rightarrow\frac{x+102}{100}+\frac{x+102}{99}+\frac{x+102}{98}-\frac{x+102}{97}-\frac{x+102}{96}-\frac{x+102}{95}\)

\(\Rightarrow\left(x+102\right)\left(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\right)=0\)

Do \(\frac{1}{100}+\frac{1}{99}+\frac{1}{98}-\frac{1}{97}-\frac{1}{96}-\frac{1}{95}\ne0\)

\(\Rightarrow x+102=0\Rightarrow x=-102\)

Vậy x = -102

c) Ta có : (x + 2) - (x + 3) = x + 2 - x - 3

= x - x + 2 - 3

= -1

mà (x + 2) - (x + 3) > 0 => không tồn tại x sao cho (x + 2) - (x + 3) > 0

d) Ta có : \(\left(x-5\right)\left(x+\frac{7}{3}\right)\ge0\)

\(\Rightarrow\orbr{\begin{cases}x\ge5\\x\ge\frac{-7}{3}\end{cases}}\)

\(\Rightarrow x\ge\frac{-7}{3}\)

Vậy \(x\ge\frac{-7}{3}\)

MH

minh hoang cong

9 tháng 8 2019

Bai 1: a) \(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\)
b) \(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
bài 2:
a) \(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\)
b) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
c)

#Toán lớp 6

2

NT

nguyen thi hien

9 tháng 8 2019

\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)

\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)

\(< =>\frac{-15+12-24}{63}\)

\(< =>\frac{-3}{7}\)

\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)

\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)

\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)

\(< =>\frac{7}{5}-\frac{4}{7}\)

\(< =>\frac{29}{35}\)

\(bai2:\)

\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)

\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)

\(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\)

\(< =>x=\frac{3}{5}:\frac{-3}{4}\)

\(< =>x=\frac{-4}{5}\)

\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)

\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)

\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)

\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)

MM

minh man

9 tháng 8 2019

Bài 1:

a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\) b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)

\(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\) \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\)

\(=\frac{3}{7}.\frac{-9}{9}\) \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)

\(=\frac{-3}{7}\) \(=\frac{7}{5}-\frac{4}{7}\)

\(=\frac{29}{35}\)

Bài 2:

a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\) b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)

\(\frac{-3}{4}x\) \(=\frac{1}{5}+\frac{4}{10}\) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)

\(\frac{-3}{4}x\) \(=\frac{3}{5}\) \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)

\(x\) \(=\frac{3}{5}:\frac{-3}{4}\) \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)

\(x\) \(=\frac{4}{-5}\) \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)

\(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\)

\(x.\frac{10}{3}=\frac{13}{12}\)

\(x=\frac{13}{12}:\frac{10}{3}\)

\(x=\frac{13}{40}\)

PD

Phạm Đức Anh

15 tháng 3 2019

Tìm số nguyên x

a) \(\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)

b)\(\frac{5}{17}+\frac{-9}{4}+\frac{-26}{31}+\frac{12}{17}+\frac{-11}{31}< \frac{x}{9}\le\frac{-3}{7}+\frac{7}{15}+\frac{4}{-7}+\frac{8}{15}\)

#Toán lớp 6

2

HQ

Huỳnh Quang Sang

15 tháng 3 2019

\(a)\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{2}{7}+\frac{-1}{4}+\frac{3}{5}+\frac{5}{7}\)

\(\Rightarrow\frac{1}{3}+\frac{1}{6}+\frac{-2}{5}+\frac{-1}{5}\le x< \frac{-3}{4}+\frac{-1}{4}+\frac{2}{7}+\frac{5}{7}+\frac{3}{5}\)

\(\Rightarrow\frac{2}{6}+\frac{1}{6}+\frac{-3}{5}\le x< -1+1+\frac{3}{5}\)

\(\Rightarrow\frac{1}{2}+\frac{-3}{5}\le x< \frac{3}{5}\)

\(\Rightarrow\frac{-1}{10}\le x< \frac{6}{10}\)

\(\Rightarrow-1\le x< 6\)

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

Bài b tương tự

PD

Phạm Đức Anh

17 tháng 3 2019

bạn ơi bạn giải câu b được ko. mk ko biết làm câu b

Tính các tổng sau:\(C=\frac{5}{7}.\frac{5}{11}+\frac{5}{7}.\frac{2}{11}-\frac{5}{7}.\frac{14}{11}\)Tìm...

DT

_Diin Thỏ_

9 tháng 4 2019

#Toán lớp 6

6

PT

Phạm Trà Giang

9 tháng 4 2019

1, Tính tổng:

\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)

\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)=\frac{5}{7}\cdot\frac{-7}{11}=\frac{-5}{11}\)

2, Tìm x:

\(x+\frac{5}{5\cdot9}+\frac{4}{9\cdot13}+\frac{4}{13\cdot17}+...+\frac{4}{41\cdot45}=\frac{-37}{45}\)

\(\Rightarrow x+\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+\frac{1}{13}-\frac{1}{17}+...+\frac{1}{41}-\frac{1}{45}=\frac{-37}{45}\)

\(\Rightarrow x+\frac{1}{5}-\frac{1}{45}=\frac{-37}{45}\Rightarrow x+\frac{9}{45}-\frac{1}{45}=\frac{-37}{45}\)

\(\Rightarrow x+\frac{8}{45}=\frac{-37}{45}\Rightarrow x=\frac{-37}{45}-\frac{8}{45}=\frac{-45}{45}=-1\)

- Các bài tìm x còn lại bạn cứ theo trình tự thực hiện phép tính mà làm nhé!

BA

Biển Ác Ma

9 tháng 4 2019

\(C=\frac{5}{7}\cdot\frac{5}{11}+\frac{5}{7}\cdot\frac{2}{11}-\frac{5}{7}\cdot\frac{14}{11}\)

\(=\frac{5}{7}\cdot\left(\frac{5}{11}+\frac{2}{11}-\frac{14}{11}\right)\)

\(=\frac{5}{7}\cdot-\frac{7}{11}\)

\(=-\frac{5}{11}\)

1) Giá trị \(x\in Z\) để \(\frac{x-5}{7-x}\) là số hữu tỉ dương. x = ? 2) Cặp số nguyên dương chẵn x; y thỏa mãn biểu thức \(\frac{x}{2}+\frac{3}{7}=\frac{5}{4}\). Vậy x = .... ; y = .... 3) Giá trị \(A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{11}}{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}+\frac{\frac{1}{4}-\frac{1}{5}+\frac{1}{7}}{\frac{3}{4}-\frac{3}{5}+\frac{3}{4}}\) ...

YD

_ Yuki _ Dễ thương _

5 tháng 2 2017

1

NL

Nguyễn Lê Phước Thịnh

11 tháng 5 2022

Bài 3:

\(A=\dfrac{2\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{3\left(\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{1\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{4}\right)}{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{4}\right)}=\dfrac{2}{3}+\dfrac{1}{3}=1\)

Đúng ghi Đ, sai ghi S vào chỗ chấm:a) \(\frac{3}{7}+\frac{1}{3}x\frac{3}{7}=\frac{16}{21}x\frac{3}{7}=\frac{16}{49}\) ......b)\(\frac{5}{9}+\frac{1}{5}x\frac{5}{6}=\frac{5}{9}x\frac{1}{6}=\frac{13}{18}\) ......c)\(\frac{13}{9}-\frac{7}{9}:\frac{2}{3}=\frac{6}{9}:\frac{3}{2}=1\) ......d)\(\frac{11}{8}-\frac{5}{8}:\frac{3}{4}=\frac{11}{8}-\frac{5}{8}x\frac{4}{3}=\frac{11}{8}-\frac{10}{24}=\frac{23}{24}\) ...

NA

Nguyễn Anh Thư

3 tháng 8 2017

#Toán lớp 4

6

MH

Mai Hồng Ngọc

3 tháng 8 2017

a Đ

b S

c S

d Đ

Đúng(1)

VT

Vũ Thị Thu Hằng

3 tháng 8 2017

a ) S

b ) Đ

c ) S

d ) Đ

k cho mk nhé

1) A= \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)b) Cho 3 so x,y,z la 3 so khac 0 thoa man dieu kien :\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)Hay tinh gia tri bieu...

NT

Nguyễn Thị Bảo Trâm

25 tháng 3 2018

3

PM

Phùng Minh Quân

25 tháng 3 2018

Bài 1 :

Ta có :

\(A=\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)

\(A=\frac{3\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)

\(A=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)

\(A=\frac{3}{5}+\frac{2}{5}\)

\(A=1\)

\(b)\) Áp dụng tính chất dãy tỉ số bằng nhau ta có :

\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}=\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)

Đo đó :

\(\frac{y+z-x}{x}=2\)\(\Rightarrow\)\(y+z=3x\)\(\left(1\right)\)

\(\frac{z+x-y}{y}=2\)\(\Rightarrow\)\(x+z=3y\)\(\left(2\right)\)

\(\frac{x+y-z}{z}=2\)\(\Rightarrow\)\(x+y=3z\)\(\left(3\right)\)

Lại có : \(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}\)

Thay (1), (2) và (3) vào \(B=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}\) ta được :

\(B=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}=\frac{8xyz}{xyz}=8\)

Vậy \(B=8\)

Chúc bạn học tốt ~

DT

Đào Thu Hoà

25 tháng 3 2018

bạn phùng minh quân câu 1 a tại sao lại rút gọn được \(\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}=\frac{3}{5}\) vậy nó không cùng nhân tử mà

câu b \(\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{\left(y-y+y\right)+\left(-x+x+x\right)+\left(z+z-z\right)}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)sao lại ra bằng 2

(mình chỉ góp ý thôi nha tại mình làm thấy nó sai sai)