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11 tháng 1 2017

Ta có :1.2+2.4+3.6+4.8+5.10/3.4+6.8+9.12+12.16+15.20=[1.2(1+4+9+...+25)]/[3.4(1+4+9+16)]

=(1.2)/(3.4)=2/12=1/6

CHÚC BẠN HỌC TỐT!!

cho tớ nhé!!!!

9 tháng 3 2019

=2+8+18+32+50/12+48+108+192+300

=110/660

=0.166666667\

18 tháng 6 2019

\(\frac{4^5\cdot9^42\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\frac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot\left(2.3\right)^9}{2^{10}\cdot3^8+\left(2.3\right)^8\cdot2^2.5}\)

\(=\frac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)

\(=\frac{2^{10}\left(3^8-3^9\right)}{2^{10}\cdot\left(3^8+3^8\cdot5\right)}=\frac{2^{10}\cdot\left(-13122\right)}{2^{10}\cdot39366}=\frac{-13122}{39366}=-\frac{1}{3}\)

25 tháng 6 2018

\(1,\\ x+\dfrac{1}{2}=-\dfrac{5}{3}\\ x=-\dfrac{5}{3}-\dfrac{1}{2}\\ x=-\dfrac{13}{6}\\ Vậyx=-\dfrac{13}{6}\)

\(2,\\ \dfrac{1}{3}-x=\dfrac{3}{5}\\ x=\dfrac{1}{3}-\dfrac{3}{5}\\ x=-\dfrac{4}{15}\\ Vậyx=-\dfrac{4}{15}\)

\(3,\\ 3-4+x=\dfrac{7}{2}\\ -1+x=\dfrac{7}{2}\\ x=\dfrac{7}{2}+1\\ x=\dfrac{9}{2}\\ Vậyx=\dfrac{9}{2}\)

\(4,\\ x-\dfrac{4}{3}=-\dfrac{7}{9}\\ x=-\dfrac{7}{9}+\dfrac{4}{3}\\ x=\dfrac{15}{27}\\ Vậyx=\dfrac{15}{27}\)

\(5,\\ x-\left(-\dfrac{7}{3}\right)=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{3}\\ x=-\dfrac{27}{18}\\ Vậyx=-\dfrac{27}{18}\)

\(6,\\ x-\dfrac{1}{5}=\dfrac{9}{10}\\ x=\dfrac{9}{10}+\dfrac{1}{5}\\ x=\dfrac{11}{10}\\ Vậyx=\dfrac{11}{10}\)

\(7,\\ x+\dfrac{5}{12}=\dfrac{3}{8}\\ x=\dfrac{3}{8}-\dfrac{5}{12}\\ x=-\dfrac{1}{24}\\ Vậyx=-\dfrac{1}{24}\)

\(8,\\ x+\dfrac{5}{4}=\dfrac{7}{6}\\ x=\dfrac{7}{6}-\dfrac{5}{4}\\ x=-\dfrac{9}{24}\\ Vậyx=-\dfrac{9}{24}\)

\(9,\\ x-\dfrac{2}{7}=\dfrac{1}{35}\\ x=\dfrac{1}{35}+\dfrac{2}{7}\\ x=\dfrac{11}{35}\\ Vậyx=\dfrac{11}{35}\\ 10,\\ x-\dfrac{1}{5}=-\dfrac{7}{10}\\ x=-\dfrac{7}{10}+\dfrac{1}{5}\\ x=-\dfrac{1}{2}\\ Vậyx=-\dfrac{1}{2}\)

18 tháng 2 2021

ahiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii hãy ấn a

nhiều quá :((

\(a,2\left(x-5\right)-3\left(x+7\right)=14\)

\(2x-10-3x-21=14\)

\(-x-31=14\)

\(-x=45\)

\(x=45\)

\(b,5\left(x-6\right)-2\left(x+3\right)=12\)

\(5x-30-2x-6=12\)

\(3x-36==12\)

\(3x=48\)

\(x=16\)

\(c,3\left(x-4\right)-\left(8-x\right)=12\)

\(3x-12-8+x=0\)

\(4x-20=0\)

\(4x=20\)

\(x=5\)

Cố nốt nha bn ! 

19 tháng 3 2020

cảm ơn, bn nha:)))

mà hình như bạn TOP 3 trả lời câu hỏi pải ko nhỉ???

5 tháng 1 2022

1/   Suy ra: -35 + 7x - 2x + 20 = 15

                5x - 15 = 15

                 5x = 30

                  x = 6

2/  Suy ra:  4x - 20 - 3x - 21 = -19

                   x - 41  = -19

                    x  = 22

3/   Suy ra:   8 - 4x + 3x - 15 = 14

                    -7 -x = 14

                      x = -7 -14

                      x = -21

4/    Suy ra:   7x - 63 - 30 + 5x = -6 + 11x

                      7x + 5x - 11x = -6 +63 + 30

                       x   =    87

5/    Suy ra:  -21x +35 + 14x - 28 = 28

                       -21x + 14x = 28 +28 - 35

                         -7x = 21

                             x= -3

6/     Suy ra:  20 - 5x + 7x - 14 = -2

                        2x   =   -8

                          x = -4

7/    Suy ra:   4x +15 -5x = 7

                      -x = -8

                        x = 8

 8/   Suy ra:   5x - 35 +30 -10x = 20

                      -5x - 5 = 20

                          -5x = 25

                            x = -5

9/    Suy ra:     35 - 7x + 5x - 10 = 15

                         -7x + 5x = 15 +10 - 35

                        -2x = -10

                          x = 5

                      

28 tháng 3 2020

Hỏi đáp ToánHỏi đáp Toán

Bài 46:

11: Ta có: \(-4\left|x-2\right|=-8\)

\(\Leftrightarrow\left|x-2\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=2\\x-2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)

Vậy: x∈{0;4}

12: Ta có: \(5\left|x+2\right|=-10\cdot\left(-2\right)\)

\(\Leftrightarrow5\left|x+2\right|=20\)

\(\Leftrightarrow\left|x+2\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-6\end{matrix}\right.\)

Vậy: x∈{-6;2}

13: Ta có: \(6\left|x-2\right|=18:\left(-3\right)\)

\(\Leftrightarrow6\left|x-2\right|=-6\)(1)

Ta có: \(\left|x-2\right|\ge0\forall x\)

\(\Rightarrow6\left|x-2\right|\ge0\forall x\)(2)

Ta có: -6<0(3)

Từ (1), (2) và (3) suy ra x∈∅

Vậy: x∈∅

14: Ta có:\(-7\left|x+4\right|=21:\left(-3\right)\)

\(\Leftrightarrow-7\left|x+4\right|=-7\)

\(\Leftrightarrow\left|x+4\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)

Vậy: x∈{-5;-3}

15: Ta có: \(4\left|x+1\right|=8\left(-2\right)-8\left(-5\right)\)

\(\Leftrightarrow4\left|x+1\right|=-16-\left(-40\right)\)

\(\Leftrightarrow4\left|x+1\right|=24\)

\(\Leftrightarrow\left|x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)

Vậy: x∈{-7;5}

16: Ta có: \(3\left|x+5\right|=-9\)(4)

Ta có: |x+5|≥0∀x

⇒3|x+5|≥0∀x(5)

Ta có: -9<0(6)

Từ (4), (5) và (6) suy ra x∈∅

Vậy: x∈∅

17: Ta có: \(-8\left|x-3\right|=24-16:2\)

\(\Leftrightarrow-8\left|x-3\right|=16\)

\(\Leftrightarrow\left|x-3\right|=-2\)

mà |x-3|≥0>-2∀x

nên x∈∅

Vậy: x∈∅

18: Ta có: \(-3\left|x+6\right|=6\cdot2-9\)

\(\Leftrightarrow-3\left|x+6\right|=3\)

\(\Leftrightarrow\left|x+6\right|=-1\)

mà |x+6|≥0>-1∀x

nên x∈∅

Vậy: x∈∅

19: Ta có: \(5-\left|x+7\right|=4\)

\(\Leftrightarrow\left|x+7\right|=1\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=-1\\x+7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=-6\end{matrix}\right.\)

Vậy: x∈{-8;-6}

20: Ta có: \(12-\left|x+8\right|=10\)

\(\Leftrightarrow\left|x+8\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=2\\x+8=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=-10\end{matrix}\right.\)

Vậy: x∈{-10;-6}

1) |x + 2| = 4

\(\Leftrightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)

2) 3 – |2x + 1| = (-5)

\(\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}\)

3) 12 + |3 – x| = 9

\(\Leftrightarrow\left|3-x\right|=-3\)(vô lí)

=>\(x=\varnothing\) 

25 tháng 2 2020

1) I x+2 I=4

\(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}}\)

2) \(3-|2x+1|=-5\)

\(\Leftrightarrow|2x+1|=8\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}}\)

3) \(12+|3-x|=9\)

\(\Leftrightarrow|3-x|=-3\)(vô lí vì I 3-x I \(\ge\)0)

21 tháng 9 2016

B= {[ 5 * (2^2)^15 * (3^2)^9 ] - [ (2^2) * 3^20 * (2^3)^9 ]} / {[ 5 * (2^9) * (2^19)*(3^19) ] - [ 7 * (2^29) * (3^3)^6 ]} 
B= {[ 5 * (2^30) * (3^18) ] - [ (3^20) * (2^29) ]} / {[ 5 * (2^28) * (3^19) ] - [ 7 * (2^29) * (3^18) ]} 
B= {[ (2^29) * (3^18) ] * [(5 * 2) - 3^2 ]} / {[ (2^28) * (3^18) ] - [(5 * 3) - (7 * 2)] } 
B= [ (2^29) * (3^18) ] / [ (2^28) * (3^18) ] 
B= [ (2^1) * (2^28) * (3^18) ] / [ (2^28) * (3^18) ] 
B = 2

dấu * là dấu nhân

12 tháng 4 2018

B= {[ 5 * (2^2)^15 * (3^2)^9 ] - [ (2^2) * 3^20 * (2^3)^9 ]} / {[ 5 * (2^9) * (2^19)*(3^19) ] - [ 7 * (2^29) * (3^3)^6 ]} 
B= {[ 5 * (2^30) * (3^18) ] - [ (3^20) * (2^29) ]} / {[ 5 * (2^28) * (3^19) ] - [ 7 * (2^29) * (3^18) ]} 
B= {[ (2^29) * (3^18) ] * [(5 * 2) - 3^2 ]} / {[ (2^28) * (3^18) ] - [(5 * 3) - (7 * 2)] } 
B= [ (2^29) * (3^18) ] / [ (2^28) * (3^18) ] 
B= [ (2^1) * (2^28) * (3^18) ] / [ (2^28) * (3^18) ] 
B = 2

dấu * là dấu nhân

27 tháng 6 2018

1/ \(x+\dfrac{1}{2}=\dfrac{-5}{3}\)

\(x=\dfrac{-5}{3}-\dfrac{1}{2}\)

\(x=\dfrac{-10}{6}-\dfrac{3}{6}\)

Vậy \(x=\dfrac{-13}{6}\)

2/\(\dfrac{1}{3}-x=\dfrac{3}{5}\)

\(-x=\dfrac{3}{5}-\dfrac{1}{3}\)

\(-x=\dfrac{9}{15}-\dfrac{5}{15}\)

\(-x=\dfrac{4}{15}\)

Vậy \(x=\dfrac{-4}{15}\)

3/ \(3-4+x=\dfrac{7}{2}\)

\(-4+x=\dfrac{7}{2}-3\)

\(-4+x=\dfrac{7}{2}-\dfrac{6}{2}\)

\(-4+x=\dfrac{1}{2}\)

\(x=\dfrac{1}{2}+4\)

\(x=\dfrac{1}{2}+\dfrac{8}{2}\)

Vậy \(x=\dfrac{9}{2}\)

4/ \(x-\dfrac{4}{3}=\dfrac{-7}{9}\)

\(x=\dfrac{-7}{9}+\dfrac{4}{3}\)

\(x=\dfrac{-7}{9}+\dfrac{12}{9}\)

Vậy \(x=\dfrac{5}{9}\)

5/ \(x-\dfrac{-7}{2}=\dfrac{5}{6}\)

\(x=\dfrac{5}{6}-\dfrac{7}{2}\)

\(x=\dfrac{5}{6}-\dfrac{21}{6}\)

Vậy \(x=\dfrac{-16}{6}=\dfrac{-8}{3}\)

6/ \(x-\dfrac{1}{5}=\dfrac{9}{10}\)

\(x=\dfrac{9}{10}+\dfrac{1}{5}\)

\(x=\dfrac{9}{10}+\dfrac{2}{10}\)

Vậy \(x=\dfrac{11}{10}\)

7/ \(x+\dfrac{5}{12}=\dfrac{3}{8}\)

\(x=\dfrac{3}{8}-\dfrac{5}{12}\)

\(x=\dfrac{9}{24}-\dfrac{10}{24}\)

Vậy \(x=\dfrac{-1}{24}\)

8/ \(x+\dfrac{5}{4}=\dfrac{7}{6}\)

\(x=\dfrac{7}{6}-\dfrac{5}{4}\)

\(x=\dfrac{14}{12}-\dfrac{15}{12}\)

Vậy \(x=\dfrac{-1}{12}\)

9/ \(x-\dfrac{2}{7}=\dfrac{1}{35}\)

\(x=\dfrac{1}{35}+\dfrac{2}{7}\)

\(x=\dfrac{1}{35}+\dfrac{10}{35}\)

Vậy \(x=\dfrac{11}{35}\)

10 /\(x-\dfrac{1}{5}=\dfrac{-7}{10}\)

\(x=\dfrac{-7}{10}+\dfrac{1}{5}\)

\(x=\dfrac{-7}{10}+\dfrac{2}{10}\)

Vậy \(x=\dfrac{-5}{10}=\dfrac{-1}{2}\)