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\(A=\left(37,1-4,5\right)-\left(-4,5+37,1\right)\)
\(=37,1-4,5+4,5-37,1\)
\(=37,1-37,1-4,5+4,5\)
\(=0.\)
\(B=-\left(315.4+275\right)+4.315-\left(10-275\right)\)
\(=-315.4-275+4.315-10+275\)
\(=-315.5+4.315-275+275-10\)
\(=0+0-10=-10.\)
\(C=-\left(\frac{3}{7}+\frac{3}{8}\right)-\left(-\frac{3}{8}+\frac{4}{7}\right)\)
\(=-\frac{3}{7}-\frac{3}{8}+\frac{3}{8}-\frac{4}{7}\)
\(=-\frac{3}{7}-\frac{4}{7}-\frac{3}{8}+\frac{3}{8}\)
\(=-1+0=-1.\)
A=37,1 - 4,5 + 4,5 - 37,1
A=(37,1 - 37,10) + ( 4,5 - 4,5 )
A = 0 + 0 = 0
B= -315.4 - 275 + 4.315 - 10 + 275
B=(-315.4 + 4.315) + ( 275-275) - 10
B= 0 + 0 - 10 = -10
C= -3/7 - 3/8 + 3/8 - 4/7
C = ( -3/7-4/7) + ( 3/8 - 3/8)
C=-7/7 + 0 = -7/7 = -1
a. = (50,9 - 50,8) * 49,1
= 1 * 49,1
= 49,1
mấy câu dưới tương tự vs cả lp 7 bài này quá dễ r mà
f, \(\dfrac{2^9.4^{10}}{8^8}=\dfrac{2^9.\left(2^2\right)^{10}}{\left(2^3\right)^8}=\dfrac{2^9.2^{20}}{2^{24}}=\dfrac{2^{29}}{2^{24}}=2^5=32\)
a: \(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{14}{25}+\dfrac{11}{25}+\dfrac{2}{7}=\dfrac{2}{7}\)
b: \(=\dfrac{3}{7}-\dfrac{5}{2}-\dfrac{3}{5}+\dfrac{4}{7}+\dfrac{3}{2}-\dfrac{2}{5}=1-1-1=-1\)
c: \(=\dfrac{4}{25}+\dfrac{7}{5}\cdot\dfrac{5}{2}-2=\dfrac{4}{25}+\dfrac{7}{2}-2=\dfrac{83}{50}\)
\(b,=-\dfrac{40}{30}-\dfrac{12}{30}-\dfrac{45}{30}=-\dfrac{97}{30}\\ c,=\left(\dfrac{4}{5}+\dfrac{7}{10}\right)+\dfrac{2}{7}=\dfrac{3}{2}+\dfrac{2}{7}=\dfrac{25}{14}\\ d,=\dfrac{2}{3}+\dfrac{7}{4}+\dfrac{1}{2}+\dfrac{3}{8}\\ =\left(\dfrac{2}{3}+\dfrac{1}{2}\right)+\left(\dfrac{7}{4}+\dfrac{3}{8}\right)=\dfrac{7}{6}+\dfrac{17}{8}=\dfrac{79}{24}\)
c: \(\dfrac{4}{5}-\dfrac{-2}{7}-\dfrac{-7}{10}\)
\(=\dfrac{56}{70}+\dfrac{20}{70}+\dfrac{49}{70}\)
\(=\dfrac{125}{70}=\dfrac{25}{14}\)
a) 1/20 - (x - 8/5) = 1/10
x - 8/5 = 1/20 - 1/10
x - 8/5 = -1/20
x = -1/20 + 8/5
x = 31/20
b) 7/4 - (x + 5/3) = -12/5
x + 5/3 = 7/4 + 12/5
x + 5/3 = 83/20
x = 83/20 - 5/3
x = 149/60
c) x - [17/2 - (-3/7 + 5/3)] = -1/3
x - (17/2 - 26/21) = -1/3
x - 305/42 = -1/3
x = -1/3 + 305/42
x = 97/14
a. \(\dfrac{3}{4}-\left|2x+1\right|=\dfrac{7}{8}\)
=> \(\left|2x+1\right|=\dfrac{3}{4}-\dfrac{7}{8}\)
=> \(\left|2x+1\right|=\dfrac{-1}{8}\)
=> \(\left\{{}\begin{matrix}2x+1=\dfrac{-1}{8}\\2x+1=\dfrac{1}{8}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=\dfrac{-9}{16}\\x=\dfrac{-7}{16}\end{matrix}\right.\)
#Yiin
b. \(2.\left|2x-3\right|=\dfrac{1}{2}\)
=> \(\left|2x-3\right|=\dfrac{1}{4}\)
=> \(\left\{{}\begin{matrix}2x-3=\dfrac{1}{4}\\2x-3=\dfrac{-1}{4}\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x=\dfrac{13}{8}\\x=\dfrac{11}{8}\end{matrix}\right.\)
a) \(=\left(13\dfrac{2}{7}+2\dfrac{5}{7}\right):\left(-\dfrac{8}{9}\right)\)
\(=16:\dfrac{-8}{9}=\dfrac{-8\cdot\left(-2\right)\cdot9}{-8}=-18\)
b)
\(=\left(\dfrac{-6}{11}\cdot\dfrac{11}{-6}\right)\cdot\dfrac{7\cdot10\cdot\left(-2\right)}{10}\)
\(=-14\)
c) \(=\dfrac{-1}{2}\cdot\dfrac{4}{3}\cdot\dfrac{-7}{2}\)
\(=\dfrac{-1\cdot2\cdot2\cdot\left(-7\right)}{2\cdot3\cdot2}=\dfrac{7}{3}\)
Bạn tính hai vế à.!? Hay tính vế thứ nhất rồi với vế thứ 2.!???
\(a,2\dfrac{1}{2}-x+\dfrac{4}{5}=\dfrac{2}{3}-\left(-\dfrac{4}{7}\right)\\ \Rightarrow\dfrac{5}{2}-x+\dfrac{4}{5}=\dfrac{26}{21}\\ \Rightarrow\dfrac{5}{2}-x=\dfrac{46}{105}\\ \Rightarrow x=\dfrac{433}{210}\\ b,-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow2x-\dfrac{4}{7}-x=\dfrac{3}{5}\\ \Rightarrow2x-x=\dfrac{41}{35}\\ \Rightarrow x=\dfrac{41}{35}\\ c,\left(\dfrac{3}{8}-\dfrac{1}{5}\right)+\left(\dfrac{5}{8}-x\right)=\dfrac{1}{5}\\ \Rightarrow\dfrac{7}{40}+\dfrac{5}{8}-x=\dfrac{1}{5}\\ \Rightarrow\dfrac{4}{5}-x=\dfrac{1}{5}\\ \Rightarrow x=\dfrac{3}{5}.\)
Áp dụng công thức bỏ dấu ngoặc:
+ có dấu trừ đằng trước-> đổi dấu tất cả các hạng tử trong ngoặc
+ có dấu cộng đằng trước-> để nguyên dấu các hạng tử trong ngoặc
\(A=\left(37,1-4,5\right)-\left(-4,5\right)+37,1\)
\(A=37,1-4,5+4,5+37,1\)
\(A=2.37,1=74,2\)
\(B=-\left(315,4+275\right)+4,315-\left(10-275\right)\)
\(B=-315,4-275+4,315-10+275\)
\(B=-315,4+4,315-10=-321,085\)
\(C=-\left(\dfrac{3}{7}+\dfrac{3}{8}\right)-\left(-\dfrac{3}{8}+\dfrac{4}{7}\right)\)
\(C=-\dfrac{3}{7}-\dfrac{3}{8}+\dfrac{3}{8}-\dfrac{4}{7}\)
\(C=-1\)
Chúc bạn học tốt!!!