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1 tháng 2 2020

\(A=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-1\frac{15}{17}+\frac{2}{3}=\frac{15}{34}+\frac{7}{21}+\frac{9}{34}-\frac{64}{34}+\frac{14}{21}=\left(\frac{15}{34}+\frac{9}{34}-\frac{64}{34}\right)+\left(\frac{7}{21}+\frac{14}{21}\right)=\frac{30}{34}+\frac{21}{21}=\frac{15}{17}+1=\frac{32}{17}\)

11 tháng 12 2019

a) 12. \(\frac{4}{9}\)+\(\frac{4}{3}\)=\(\frac{16}{3}\)+\(\frac{4}{3}\)=\(\frac{20}{3}\)

b) (\(\frac{-5}{7}\)) . (12,5+1,5)= (\(\frac{-5}{7}\)).14=-10

a) \(12.\left(-\frac{2}{3}\right)^2+\frac{4}{3}=12.\frac{4}{9}+\frac{4}{3}=\frac{16}{3}+\frac{4}{3}=\frac{20}{3}\)

b) \(12,5.\left(-\frac{5}{7}\right)+1,5.\left(-\frac{5}{7}\right)=-\frac{5}{7}.\left(12,5+1,5\right)=-\frac{5}{7}.14=-10\)

c) \(1:\left(\frac{2}{3}-\frac{3}{4}\right)^2=1:\left(-\frac{1}{12}\right)^2=1:\frac{1}{144}=1.144=144\)

d) \(15.\left(-\frac{2}{3}\right)^2-\frac{7}{3}=15.\frac{4}{9}-\frac{7}{3}=\frac{20}{3}-\frac{7}{3}=\frac{13}{3}\)

e) \(\frac{1}{2}\sqrt{64}-\sqrt{\frac{4}{25}}+\left(-1\right)^{2007}=\frac{1}{2}.8-\frac{2}{5}+\left(-1\right)=4-\frac{2}{5}-1=\frac{13}{5}\)

26 tháng 2 2020

Bài 1:

a) \(0,5-\frac{5}{41}+\frac{1}{2}-\frac{36}{41}\)

\(=\frac{1}{2}-\frac{5}{41}+\frac{1}{2}-\frac{36}{41}\)

\(=\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{5}{41}+\frac{36}{41}\right)\)

\(=1-1\)

\(=0.\)

b) \(\left(-\frac{2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(-\frac{1}{3}+\frac{4}{7}\right):\frac{4}{5}\)

\(=-\frac{2}{3}+\frac{3}{7}:\frac{4}{5}-\frac{1}{3}+\frac{4}{7}:\frac{4}{5}\)

\(=\left[\left(-\frac{2}{3}\right)-\frac{1}{3}\right]+\left(\frac{3}{7}+\frac{4}{7}\right):\frac{4}{5}\)

\(=\left(-1\right)+1:\frac{4}{5}\)

\(=\left(-1\right)+\frac{5}{4}\)

\(=\frac{1}{4}.\)

c) \(\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+3.\sqrt{49}}\)

\(=\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+3.7}\)

\(=\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+21}\)

\(=\left(-\frac{3}{4}\right).\sqrt{\frac{205}{9}}\)

\(=\left(-\frac{3}{4}\right).\frac{\sqrt{205}}{3}\)

\(=-\frac{\sqrt{205}}{4}.\)

d) \(\left(-\frac{1}{3}\right)^2.\frac{4}{11}+1\frac{5}{11}.\left(\frac{1}{3}\right)^2\)

\(=\frac{1}{9}.\frac{4}{11}+\frac{16}{11}.\frac{1}{9}\)

\(=\frac{1}{9}.\left(\frac{4}{11}+\frac{16}{11}\right)\)

\(=\frac{1}{9}.\frac{20}{11}\)

\(=\frac{20}{99}.\)

Chúc bạn học tốt!

26 tháng 2 2020

cảm ơn bạn

AH
Akai Haruma
Giáo viên
1 tháng 1 2020

Bài 1:

a)

\((\frac{3}{5})^2-[\frac{1}{3}:3-\sqrt{16}.(\frac{1}{2})^2]-(10.12-2014)^0\)

\(=\frac{9}{25}-(\frac{1}{9}-1)-1\)

\(=\frac{9}{25}-\frac{1}{9}=\frac{56}{225}\)

b)

\(|-\frac{100}{123}|:(\frac{3}{4}+\frac{7}{12})+\frac{23}{123}:(\frac{9}{5}-\frac{7}{15})\)

\(=\frac{100}{123}:\frac{4}{3}+\frac{23}{123}:\frac{4}{3}=(\frac{100}{123}+\frac{23}{123}):\frac{4}{3}=1:\frac{4}{3}=\frac{3}{4}\)

c)

\(\frac{(-5)^{32}.20^{43}}{(-8)^{29}.125^{25}}=\frac{5^{32}.(2^2.5)^{43}}{(-2)^{3.29}.(5^3)^{25}}=\frac{5^{32}.2^{86}.5^{43}}{-2^{87}.5^{75}}\)

\(=\frac{5^{32+43}.2^{86}}{-2^{87}.5^{75}}=\frac{5^{75}.2^{86}}{-2^{87}.5^{75}}=-\frac{1}{2}\)

AH
Akai Haruma
Giáo viên
1 tháng 1 2020

Bài 2:

a)

\(\frac{2}{3}-(\frac{3}{4}-x)=\sqrt{\frac{1}{9}}=\frac{1}{3}\)

\(\frac{3}{4}-x=\frac{2}{3}-\frac{1}{3}=\frac{1}{3}\)

\(x=\frac{3}{4}-\frac{1}{3}=\frac{5}{12}\)

b)

\((\frac{1}{2}-x)^2=(-2)^2=2^2\)

\(\Rightarrow \left[\begin{matrix} \frac{1}{2}-x=-2\\ \frac{1}{2}-x=2\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{5}{2}\\ x=\frac{-3}{2}\end{matrix}\right.\)

c)

\(|3x+\frac{1}{2}|-\frac{2}{3}=1\)

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

\(\Rightarrow \left[\begin{matrix} 3x+\frac{1}{2}=\frac{5}{3}\\ 3x+\frac{1}{2}=-\frac{5}{3}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{7}{18}\\ x=\frac{-13}{18}\end{matrix}\right.\)

d)

\(3^{2x+1}=81=3^4\)

\(\Rightarrow 2x+1=4\Rightarrow x=\frac{3}{2}\)

15 tháng 3 2020

Ta có : \(\left(2^2:\frac{4}{3}-\frac{1}{2}\right).\frac{6}{5}-17\)

=\(=\left(4.\frac{3}{4}-\frac{1}{2}\right).\frac{6}{5}-17\)

\(=\frac{5}{2}.\frac{6}{5}-17\)

\(=3-17=-14\)

Tụi quá mới lớp 5 thui

5 tháng 12 2015

câu  nào bạn không làm đc

5 tháng 12 2015

\(\frac{3}{4}+\frac{1}{4}:\left(-\frac{2}{3}\right)-\left(-5\right)\)

\(=\frac{3}{4}+\frac{1}{4}.\left(-\frac{3}{2}\right)+5\)

\(=\frac{3}{4}-\frac{3}{8}+5\)

\(=\frac{3}{8}+5=\frac{43}{8}\)

\(12.\left(\frac{2}{5}-\frac{5}{6}\right)^2=12.\left(-\frac{13}{30}\right)^2=12.\frac{169}{900}=\frac{169}{75}\)

\(\left(-2\right)^2+\sqrt{36}-\sqrt{9}+\sqrt{25}=4+6-3+5=12\)

\(\left(9\frac{3}{4}:3.4.2\frac{7}{34}\right):\left(-1\frac{9}{16}\right)=\left(\frac{39}{4}:3.4.\frac{75}{34}\right):\left(-\frac{25}{16}\right)=\frac{975}{34}.\left(-\frac{16}{25}\right)=-\frac{312}{17}\)

\(\frac{\sqrt{3^2}+\sqrt{39^2}}{\sqrt{91^2}-\sqrt{\left(-7\right)^2}}=\frac{3+39}{91-7}=\frac{42}{84}=\frac{1}{2}\)

25 tháng 10 2019

\(a)=\frac{7}{25}+\frac{4}{13}-\frac{5}{2}+\frac{18}{25}-\frac{17}{13}\)

\(=1-1-\frac{5}{2}\)

\(=-\frac{5}{2}\)

25 tháng 10 2019

cái này bạn bấm máy tính là ra mà 

Bài 1:...
Đọc tiếp

Bài 1: Tính

a. \(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)+\left(1+\frac{1}{4\cdot6}\right).....\left(1+\frac{1}{99\cdot101}\right)\)

b. \(\left[\sqrt{0,64}+\sqrt{0,0001}-\sqrt{\left(-0,5\right)^2}\right]\div\left[3\cdot\sqrt{\left(0,04\right)^2}-\sqrt{\left(-2\right)^4}\right]\)

c. \(\frac{5.4^{15}\cdot9^9-4.3^{20}\cdot8^9}{5\cdot2^9\cdot6^{19}-7\cdot2^{29}\cdot27^6}-\frac{2^{19}\cdot6^{15}-7\cdot6^{10}\cdot2^{20}\cdot3^6}{9\cdot6^{19}\cdot2^9-4\cdot3^{17}\cdot2^{26}}+0,\left(6\right)\)

Bài 2: Tìm x, y, z biết :
a. \(\left(x-10\right)^{1+x}=\left(x-10\right)^{x+2009}\left(x\in Z\right)\)

b. \(\left|x-2007\right|+\left|x-2008\right|+\left|y-2009\right|+\left|x-2010\right|=3\left(x,y\in N\right)\) 

c. \(25-y^2=8\left(x-2009\right)^2\left(x,y\in Z\right)\)

d. \(2008\left(x-4\right)^2+2009\left|x^2-16\right|+\left(y+1\right)^2\le0\)

e. \(2x=3y\) ; \(4z=5x\) và \(3y^2-z^2=-33\)

Bài 3: Chứng minh rằng

a. \(1-\frac{1}{2^2}-\frac{1}{3^2}-\frac{1}{4^2}-...-\frac{1}{2009^2}>\frac{1}{2009}\)

b. \(\left[75\cdot\left(4^{2008}+4^{2007}+4^{2006}+...+4+1\right)+25\right]⋮100\)

Bài 4: 

a. Tìm giá trị nhỏ nhất của biểu thức : \(M=\left(x^2+2\right)+\left|x+y-2009\right|+2005\)

b. So sánh: \(31^{11}\) và \(\left(-17\right)^{14}\)

c. So sánh: \(\left(\frac{9}{11}-0,81\right)^{2012}\) và \(\frac{1}{10^{4024}}\)

1

Bài 1 :\(a,=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}...\frac{100^2}{99.101}\)

           \(=\frac{2.3.4...100}{1.2.3...99}.\frac{2.3.4...100}{3.4...101}\)

          \(=100.\frac{2}{101}=\frac{200}{101}\)

\(b,\left(\sqrt{1\frac{9}{16}-\sqrt{\frac{9}{16}}}\right):5\)

\(=\left(\sqrt{\frac{25}{16}-\frac{3}{4}}\right):5\)

\(=\sqrt{\frac{13}{16}}:5\)

\(=\frac{\sqrt{13}}{4}:5\)

\(=\frac{\sqrt{13}}{20}\)