Rút gọn
\(A=\frac{21.99+21}{35.201-35}\)
\(B=\frac{\left(-2\right)^{36}.3^{17}}{3^{18}.2^{35}}\)
\(C=\frac{7^{100}-7^{99}}{7^{98}-7^{97}}\)
Tính hợp lí:
a)\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{127}\)\(-\left(+\frac{7}{18}\right)+\frac{4}{35}-\left(\frac{-2}{7}\right)\)
b)\(\frac{3}{5}+\frac{3}{11}-\left(\frac{-3}{7}\right)+\)\(\left(\frac{2}{97}\right)-\frac{1}{35}-\frac{3}{4}+\left(\frac{-23}{44}\right)\)
Kiểm tra bài : Nhân, chia số hữu tỉ
Thực hiện phép tính :
(1) \(-\frac{3}{2}.\frac{7}{10}=\frac{-3.7}{2.10}=\frac{-21}{20}\)
(2) \(\frac{-5}{3}.\frac{6}{11}=\frac{-5.6}{3.11}=\frac{-30}{33}\)
(3) \(2\frac{1}{3}.\left(-1\frac{2}{3}\right)=\frac{7}{3}.\left(-\frac{5}{3}\right)=\frac{7.\left(-5\right)}{3.3}=-\frac{35}{9}\)
(4) \(\frac{9}{10}:\left(-\frac{15}{11}\right)=\frac{9}{10}.\left(\frac{-11}{15}\right)=\frac{9.\left(-11\right)}{10.15}=-\frac{99}{150}=-\frac{33}{50}\)
(5) \(\left(-1\right):\frac{3}{8}=\frac{\left(-1\right).8}{3}=-\frac{8}{3}\)
(6) \(\frac{1}{2}.\left(-\frac{5}{4}\right).\frac{8}{7}=\frac{1.\left(-5\right)}{2.4}.\frac{8}{7}=-\frac{5}{8}.\frac{8}{7}=-\frac{5.8}{8.7}=-\frac{5}{7}\)
(7) \(\frac{-9}{2}.\frac{2}{18}.\frac{1}{7}=\left(-\frac{9}{2}.\frac{2}{18}\right).\frac{1}{7}=\left(-\frac{9.2}{2.18}\right).\frac{1}{7}=-\frac{18}{36}.\frac{1}{7}=-\frac{18.1}{36.7}=-\frac{1}{14}\)
(8) \(\left(\frac{9}{2}-\frac{1}{3}\right).\frac{6}{17}=\left(\frac{27}{6}-\frac{2}{6}\right).\frac{6}{17}=\frac{27-2}{6}.\frac{6}{17}=\frac{25}{6}.\frac{6}{17}=\frac{25.6}{6.17}=\frac{25}{17}\)
(9) \(\left(-\frac{12}{13}:\frac{36}{39}\right).\frac{3}{5}=\left(-\frac{12}{13}.\frac{39}{36}\right).\frac{3}{5}=\left(\frac{-12.39}{13.36}\right).\frac{3}{5}=-\frac{1.3}{5}=-\frac{3}{5}\)
(10) \(\left(-\frac{3}{7}+\frac{7}{9}\right):\frac{4}{7}+\left(-\frac{4}{7}+\frac{2}{9}\right):\frac{4}{7}=\left(\left(-\frac{3}{7}+\frac{7}{9}\right)+\left(-\frac{4}{7}+\frac{2}{9}\right)\right):\frac{4}{7}\)
\(=\left(\left(-\frac{27}{63}+\frac{49}{63}\right)+\left(-\frac{36}{63}+\frac{14}{63}\right)\right):\frac{4}{7}=\left(\left(-\frac{27+49}{63}\right)+\left(\frac{-36+14}{63}\right)\right):\frac{4}{7}\)
\(=\left(\left(\frac{22}{63}\right)+\left(-\frac{22}{63}\right)\right):\frac{4}{7}\)
\(=\frac{22+\left(-22\right)}{63}:\frac{4}{7}=\frac{0}{63}:\frac{4}{7}=0\)
Mình đăng các bài toán này lên thứ nhất là để kiểm tra năng lực thứ hai các bạn có thể xem đây và rút ra lời giải cho các bài khác và nếu mình sai chỗ nào các bạn chỉ mình sẽ chỉnh
Bài 1 Thưc hiện phép tính ( tính nhanh nếu có thể)
a)\(\frac{-1}{24}-\left[\frac{1}{4}-\left(\frac{1}{2}-\frac{7}{8}\right)\right]\)
b)\(\left(\frac{5}{7}-\frac{7}{5}\right)-\left[\frac{1}{2}-\left(\frac{-2}{7}-\frac{1}{10}\right)\right]\)
C)\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{17}-\left(\frac{-2}{7}\right)+\frac{4}{35}-\frac{7}{18}\)
d)\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
Thực hiện phép tính một cách hợp lí
a/ \(\frac{1}{3}-\left(\frac{-3}{4}\right)-\left(\frac{-3}{5}\right)+\frac{1}{57}-\frac{1}{36}+\frac{1}{15}+\left(\frac{-2}{9}\right)\)
b/ \(-\frac{-1}{3}-\left(-\frac{3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{127}-\frac{7}{18}+\frac{4}{35}-\left(\frac{-2}{7}\right)\)
1)Tính nhanh giá trị biểu thức
A=\(\frac{1}{3}-\frac{3}{4}-\left(\frac{-3}{5}\right)+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
B=\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{131}-\left(\frac{-2}{7}\right)+\frac{4}{35}-\frac{7}{18}\)
Câu a) bạn tham khảo tại đây nhé: Câu hỏi của Hằng Thanh.
Chúc bạn học tốt!
Tính giá trị biểu thức
\(1.A=\frac{1}{5}+\frac{3}{17}-\frac{4}{3}+\left(\frac{4}{5}-\frac{3}{17}+\frac{1}{3}\right)-\frac{1}{7}+\left[\frac{-14}{30}\right]\)
\(2.B=\left(\frac{5}{8}-\frac{4}{12}+\frac{3}{2}\right)-\left(\frac{5}{8}+\frac{9}{13}\right)-\left[\frac{-3}{2}\right]+\frac{7}{-15}\)
\(3.C=\frac{5}{18}+\frac{8}{19}-\frac{7}{21}+\left(\frac{-10}{36}+\frac{11}{19}+\frac{1}{3}\right)-\frac{5}{8}\)
\(4.D=\frac{1}{9}-\left[\frac{-5}{23}\right]-\left(\frac{-5}{23}+\frac{1}{9}+\frac{25}{7}\right)+\frac{50}{14}-\frac{7}{30}\)
\(5.E=\frac{1}{13}+\left(\frac{-5}{18}-\frac{1}{13}+\frac{12}{17}\right)+\left(\frac{12}{17}+\frac{5}{18}+\frac{7}{5}\right)\)
\(6.F=\frac{15}{14}-\left(\frac{17}{23}-\frac{80}{87}+\frac{5}{4}\right)+\left(\frac{12}{17}-\frac{15}{14}+\frac{1}{4}\right)\)
\(7.G=\frac{1}{25}-\frac{4}{27}+\left(\frac{-23}{27}+\frac{-1}{25}-\frac{5}{43}\right)+\frac{5}{43}-\frac{4}{7}\)
\(8.H=\frac{4}{15}-\frac{23}{28}-\left(\frac{-23}{28}+\frac{-11}{15}-\frac{29}{27}\right)-\frac{2}{27}\)
\(9.K=\frac{1}{16}-\frac{5}{21}+\left(\frac{-1}{16}+\frac{-3}{5}-\frac{-5}{21}\right)+\frac{-2}{5}+\frac{3}{4}\)
\(10.L=\frac{7}{12}+\frac{15}{14}-\left(\frac{14}{22}+\frac{-1}{14}+\frac{5}{21}\right)-\frac{-5}{21}+\frac{3}{5}\)
yutyugubhujyikiu
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)\(\left(x-5\right)\left(x+\frac{7}{3}\right)\ge0\)
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) \(\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(3+\dfrac{x+2}{100}+\dfrac{x+3}{99}+\dfrac{x+4}{98}=3+\dfrac{x+5}{97}+\dfrac{x+6}{96}+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(1+\dfrac{x+2}{100}+1+\dfrac{x+3}{99}+1+\dfrac{x+4}{98}=1+\dfrac{x+5}{97}+1+\dfrac{x+6}{96}+1+\dfrac{x+7}{95}\)
\(\Rightarrow\)\(\dfrac{100}{100}+\dfrac{x+2}{100}+\dfrac{99}{99}+\dfrac{x+3}{99}+\dfrac{98}{98}+\dfrac{x+4}{98}=\dfrac{97}{97}+\dfrac{x+5}{97}+\dfrac{96}{96}+\dfrac{x+6}{96}+\dfrac{95}{95}+\dfrac{x+7}{95}\)\(\Rightarrow\)\(\dfrac{x+102}{100}+\dfrac{x+102}{99}+\dfrac{x+102}{98}=\dfrac{x+102}{97}+\dfrac{x+102}{96}+\dfrac{x+102}{95}\)
\(\Rightarrow\)\(\left(x+102\right)\left(\dfrac{1}{100}+\dfrac{1}{99}+\dfrac{1}{98}\right)=\left(x+102\right)\left(\dfrac{1}{97}+\dfrac{1}{96}+\dfrac{1}{95}\right)\)
\(\Rightarrow\)\(x+102=0\)
\(\Rightarrow x=-102\)
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: \(\left[{}\begin{matrix}x-5=0\\x+\dfrac{7}{3}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-7}{3}\end{matrix}\right.\)
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)\(\left(x-5\right)\left(x+\frac{7}{3}\right)\ge0\)
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 \(x\ge\frac{-7}{3}\)
ĐỀ: TÌM X
\(a,\frac{7}{5}:x+\frac{3}{2}=\frac{16}{3}\)
\(b,x:\frac{1}{5}+\frac{1}{7}=\frac{3}{5}.\frac{18}{21}\)
\(c,x-1\frac{1}{3}:2=\frac{5}{7}\)
\(d,x+2\frac{3}{5}.\frac{1}{6}=\frac{35}{36}\)
\(e,\left(x+\frac{3}{2}\right):2=\frac{7}{10}+\frac{1}{5}\)
ai trả lời được mik tik cho
a, 7/5 : x + 3/2 = 16/3
7/5 : x = 16/3 - 3/2
7/5 : x = 23/6
x = 7/5 : 23/6
x = 42/115
b, x : 1/5 + 1/7 = 3/5 . 18/21
x : 1/5 + 1/7 = 18/35
x : 1/5 = 18/35 - 1/7
x : 1/5 = 13/35
x = 13/35 . 1/5
x = 13/175
c, x - 1 và 1/3 : 2 = 5/7
x - 4/3 : 2 = 5/7
x - 4/3 = 5/7 . 2
x - 4/3 = 10/7
x = 10/7 + 4/3
x = 58/21
d, x + 2 và 3/5 . 1/6 = 35/36
x + 13/5 . 1/6 = 35/36
x + 13/5 = 35/36 : 1/6
x + 13/5 = 35/6
x = 35/6 - 13/5
x = 97/30
e, ( x + 3/2 ) : 2 = 7/10 + 1/5
( x + 3/2 ) : 2 = 9/10
x + 3/2 = 9/10 . 2
x + 3/2 = 9/5
x = 9/5 - 3/2
x = 3/10