\(\dfrac{-5}{11}\)+(\(\dfrac{-6}{11}\)+1)
Mong các bạn giúp
thank you m.n
a)\(\dfrac{-1}{4}.13\dfrac{9}{11}-0,25.6\dfrac{2}{11}\)
b)\(\dfrac{4}{9}:\left(\dfrac{-1}{7}\right)+6\dfrac{5}{9}:\left(\dfrac{-1}{7}\right)\)
Các bạn giúp mik nha mik sẽ tick và kết bạn. Thank you
a)\(\dfrac{-1}{4}\cdot13\dfrac{9}{11}-0,25\cdot6\dfrac{2}{11}\)
\(=\dfrac{-1}{4}\cdot\dfrac{152}{11}-\dfrac{1}{4}\cdot\dfrac{68}{11}\)
=\(\dfrac{1}{4}\cdot\left(\dfrac{-152}{11}-\dfrac{68}{11}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{-220}{11}=-5\)
b)\(\dfrac{4}{9}:\left(\dfrac{-1}{7}\right)+6\dfrac{5}{9}:\left(\dfrac{-1}{7}\right)\)
\(=\dfrac{4}{9}\cdot-7+\dfrac{59}{9}\cdot-7\)
\(=-7\cdot\left(\dfrac{4}{9}+\dfrac{59}{9}\right)\)\(=-7\cdot7=-49\)
Mong các bạn trả lời nhanh !
B=\(\dfrac{11}{17}+(-\dfrac{8}{19})+(-\dfrac{3}{4})+\dfrac{6}{17}-\dfrac{30}{19}\)
`B=11/17+ (-8/19) +(-3/4) + 6/17 - 30/19`
`B=(11/17 +6/17) + (-8/19 -30/19) + (-3/4)`
`B= 17/17+(-38/19)+(-3/4)`
`B= 1 +(-2) +(-3/4)`
`B= -1 +(-3/4)`
`B=-7/4`
cho biểu thức P=\(\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{ }x}\)
a/ Rút gọn P
b/Tính P với x = 14 -\(6\sqrt{5}\)
mong mọi người giúp thank you
\(a,P=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\left(x\ge0;x\ne9\right)\\ P=\dfrac{x\sqrt{x}-3-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\\ P=\dfrac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\\ P=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\\ P=\dfrac{\left(x+8\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x+8}{\sqrt{x}+1}\)
\(b,x=14-6\sqrt{5}=\left(3-\sqrt{5}\right)^2\)
Thay vào P:
\(P=\dfrac{14-6\sqrt{5}+8}{\sqrt{\left(3-\sqrt{5}\right)^2}+1}=\dfrac{22-6\sqrt{5}}{4-\sqrt{5}}=\dfrac{\left(4+\sqrt{5}\right)\left(22-6\sqrt{5}\right)}{11}=\dfrac{55-2\sqrt{5}}{11}\)
a) \(P=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\left(đk:x\ge0,x\ne9\right)\)
\(=\dfrac{x\sqrt{x}-3-2\left(\sqrt{x}-3\right)^2-\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x\sqrt{x}-3-2x+12\sqrt{x}-18-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3x+x\sqrt{x}+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x\left(\sqrt{x}-3\right)+8\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(x+8\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x+8}{\sqrt{x}+1}\)
b) \(P=\dfrac{x+8}{\sqrt{x}+1}=\dfrac{14-6\sqrt{5}+8}{\sqrt{14-6\sqrt{5}}+1}=\dfrac{22-6\sqrt{5}}{\sqrt{\left(3-\sqrt{5}\right)^2}+1}=\dfrac{22-6\sqrt{5}}{3-\sqrt{5}+1}=\dfrac{22-6\sqrt{5}}{4-\sqrt{5}}\)
Thực hiện phép tính
a. \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}-\left(\dfrac{1}{2}-\dfrac{7}{8}\right)\right\}\)
b.\(\left(8\dfrac{5}{11}+3\dfrac{5}{8}\right)-3\dfrac{5}{11}\)
c.\(\dfrac{4}{9}:\left(\dfrac{-1}{7}\right)+6\dfrac{5}{9}:\left(\dfrac{-1}{7}\right)\)
d. \(\left(\dfrac{-1}{5}+\dfrac{3}{7}\right):\dfrac{2}{11}+\left(\dfrac{-4}{5}+\dfrac{4}{7}\right):\dfrac{2}{11}\)
M.n giúp mk nha
a. = \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}-\dfrac{-3}{8}\right\}\)
= \(\dfrac{-1}{24}-\left\{\dfrac{1}{4}+\dfrac{3}{8}\right\}\)
= \(\dfrac{-1}{24}-\dfrac{5}{8}\)
= \(\dfrac{-2}{3}\)
b. = \(12\dfrac{7}{88}-3\dfrac{5}{11}\)
= \(8\dfrac{5}{8}\)
c. = \(\dfrac{-28}{9}+\dfrac{-413}{9}\)
= \(-49\)
d. = \(\dfrac{8}{35}:\dfrac{2}{11}+\dfrac{-8}{35}:\dfrac{2}{11}\)
= \(\dfrac{2}{11}:\left(\dfrac{8}{35}+\dfrac{-8}{35}\right)\)
= 0
\(a\dfrac{-2}{3}.\dfrac{4}{5}+\dfrac{1}{5}:\dfrac{9}{11}b\left(-6,2:3,7\right):0,2\)
MONG MỌI NGƯỜI, ANH CHỊ, THẦY CÔ GIÚP ĐỠ
Giải:
a) -2/3.4/5+2/5:9/11
=-8/15+22/45
=-2/45
b) (-6,2:3,7):0,2
=-62/37:0,2
=-310/37
Chúc bạn học tốt!
a)-2/3.4/5+1/5:9/11
=-8/15+11/45
=-13/45
ok nhé!
ok giúp t thêm vài câu nữa đi :')
Tính hợp lý \(\dfrac{2}{11}\)-\(\dfrac{3}{8}\)+\(\dfrac{4}{11}\)-\(\dfrac{6}{11}\)-\(\dfrac{5}{8}\)
Số x thoả mãn \(\dfrac{1}{4}\)+\(\dfrac{x}{12}\)=\(\dfrac{8}{12}\)
Tìm x biết \(\dfrac{1}{2}\)-(x-\(\dfrac{5}{11}\))=\(\dfrac{-3}{4}\)
An đọc 1 quyển sách trong 3 ngày. Ngày thứ nhất An đọc đc \(\dfrac{1}{11}\) quyển sách,ngày thứ hai An đọc đc \(\dfrac{8}{11}\) quyển sách.Hỏi trong 2 ngày An đọc đc bao nhiêu phần quyển sách?
Bài 1 :
\(=\dfrac{2}{11}+\dfrac{4}{11}-\dfrac{6}{11}-\dfrac{3}{8}-\dfrac{5}{8}=0-1=-1\)
Bài 2 :
\(\Rightarrow3+x=8\Leftrightarrow x=5\)
Bài 3 :
\(\Leftrightarrow x-\dfrac{5}{11}=\dfrac{5}{4}\Leftrightarrow x=\dfrac{35}{44}\)
Bài 4 :
Trong 2 ngày An đọc được số quyên phần quyên sách
\(\dfrac{1}{11}+\dfrac{8}{11}=\dfrac{9}{11}\)( quyển sách )
đs : 9/11 quyển sách
Cho các số dương x, y, z thỏa mãn điều kiện \(x^2+y^2+z^2=1\).CM \(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}\ge\dfrac{1}{3}\)
mong mọi nguòi giúp thank you
Ta có: \(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}=\dfrac{x^4}{xy+2zx}+\dfrac{y^4}{yz+2xy}+\dfrac{z^4}{zx+2yz}\)
\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{xy+2zx+yz+2xy+zx+2yz}=\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\)
Mà ta lại có: \(xy+yz+zx\le x^2+y^2+z^2\)
\(\Rightarrow\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1^2}{3.1}=\dfrac{1}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{1}{\sqrt{3}}\)
cho các phân số sau
\(\dfrac{2}{12}\) \(\dfrac{8}{25}\)
\(\dfrac{5}{44}\) \(\dfrac{11}{40}\)
\(\dfrac{12}{9}\)
Tìm Thừa số nguyên tố khác nhau
Mọi người giúp mình với mình đăng lần này lần 2 rùi. Mong đc mọi người giúp đỡ
Bài 1: Thực hiện phép tính:
\(A=\left(-\dfrac{1}{125}\right)^{11}:\left(\dfrac{1}{5}\right)^{32}\)
\(B=1+\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2+....+\left(\dfrac{1}{3}\right)^{2018}\)
\(C=\dfrac{16^3\cdot3^{10}+120\cdot6^9}{4^6\cdot3^{12}+6^{11}}\)
\(D=\left(\dfrac{0.4-\dfrac{2}{9}+\dfrac{2}{11}}{1.4-\dfrac{7}{9}+\dfrac{7}{11}}-\dfrac{\dfrac{1}{3}-0.25+\dfrac{1}{5}}{1\dfrac{1}{6}-0.875+0.7}\right):\dfrac{2017}{2018}\)
\(E=\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\)
\(G=\dfrac{\left(\dfrac{2}{5}\right)^7\cdot5^7+\left(2\dfrac{1}{4}\right)^3:\left(\dfrac{3}{16}\right)^3}{512+2^7\cdot5^2}:\dfrac{\left(\dfrac{1}{2}\right)^0}{\left(-1\right)^{2017}}\)
Mn ơi giúp e với ........ Em đang cần gấp giúp e với nha!!
Thank you mn nhiều nhiều.....
Bài 1:
a: \(A=\left(-\dfrac{1}{5}\right)^{33}:\left(-\dfrac{1}{5}\right)^{32}=\dfrac{-1}{5}\)
c: \(C=\dfrac{2^{12}\cdot3^{10}+3^9\cdot2^9\cdot2^3\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)