\(B=\frac{x}{y}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{\frac{-2}{12}-\frac{10}{24}+\frac{14}{39}}\) (x=0,5;y=3)
Những câu hỏi liên quan
\(\frac{\frac{-5}{6}}{\frac{-10}{11}}+\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{\frac{-2}{12}-\frac{10}{24}+\frac{14}{29}}\)
Bài 1 :Thực hiện phép tính :
a) M (frac{-6}{13}+frac{15}{26}-frac{47}{39}-frac{1}{78}) : (99frac{17}{65}-100frac{5}{52}+frac{1}{130})
b) N frac{(frac{3}{5}-0,435+frac{1}{200}):left(-0,04right)}{30,75+frac{1}{12}+3frac{1}{6}}
c) P (frac{-5}{6}:frac{-10}{11})+frac{frac{1}{4}+frac{5}{8}-frac{7}{13}}{frac{-2}{12}-frac{10}{24}+frac{14}{39}}
Bài 2 : Thực hiện phép tính :V
a) P frac{frac{1}{5}-frac{1}{9}+frac{1}{13}}{frac{9}{5}-1+frac{9}{13}}+frac{frac{10}{7}-frac{10}{11}-frac{10}{17}}{frac{12}{...
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Bài 1 :Thực hiện phép tính :
a) M =(\(\frac{-6}{13}+\frac{15}{26}-\frac{47}{39}-\frac{1}{78}\)) : (\(99\frac{17}{65}-100\frac{5}{52}+\frac{1}{130}\))
b) N = \(\frac{(\frac{3}{5}-0,435+\frac{1}{200}):\left(-0,04\right)}{30,75+\frac{1}{12}+3\frac{1}{6}}\)
c) P = (\(\frac{-5}{6}:\frac{-10}{11}\))+\(\frac{\frac{1}{4}+\frac{5}{8}-\frac{7}{13}}{\frac{-2}{12}-\frac{10}{24}+\frac{14}{39}}\)
Bài 2 : Thực hiện phép tính :V
a) P =\(\frac{\frac{1}{5}-\frac{1}{9}+\frac{1}{13}}{\frac{9}{5}-1+\frac{9}{13}}+\frac{\frac{10}{7}-\frac{10}{11}-\frac{10}{17}}{\frac{12}{7}-\frac{12}{11}-\frac{12}{17}}\)
b) Q = \(\frac{\frac{1}{14}-\frac{1}{30}-\frac{1}{46}}{\frac{2}{35}-\frac{2}{75}-\frac{2}{115}}:\frac{\frac{3}{8}-\frac{15}{17}+\frac{30}{31}}{\frac{1}{6}-\frac{20}{51}+\frac{40}{93}}\)
có rất nhiều câu dễ ở trong đề sao bạn Ko thử làm đi rồi câu nào khó lại hỏi
1) \(\frac{24}{-12}=\frac{x}{5}=\frac{-y}{3}\)Tìm x và y
2) \(\frac{1}{3}+\frac{-2}{5}+\frac{1}{6}+\frac{-5}{25}\le\frac{x}{10}< \frac{-3}{4}+\frac{4}{14}+\frac{-2}{8}+\frac{-3}{5}+\frac{5}{7}\)Tìm x
3) \(\frac{8.x+18}{2.x+6}\)Tìm x
1. Tìm x:
a;\(\frac{x+1}{3}=\frac{x-2}{4}\)
b;\(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
c;\(\frac{x+3}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)
a) \(\frac{x+1}{3}=\frac{x-2}{4}\)
=> (x+1).4 = (x - 2) . 3
=> 4x + 4 = 3x - 6
=> 4x - 3x = - 6 - 4
=> x = - 10
b) \(\frac{x-6}{7}+\frac{x-7}{8}+\frac{x-8}{9}=\frac{x-9}{10}+\frac{x-10}{11}+\frac{x-11}{12}\)
\(\Rightarrow\left(\frac{x-6}{7}+1\right)+\left(\frac{x-7}{8}+1\right)+\left(\frac{x-8}{9}+1\right)=\left(\frac{x-9}{10}+1\right)+\left(\frac{x-10}{11}+1\right)+\left(\frac{x-11}{12}+1\right)\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}=\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}\)
\(\Rightarrow\frac{x+1}{7}+\frac{x+1}{8}+\frac{x+1}{9}-\frac{x+1}{10}-\frac{x+1}{11}-\frac{x+1}{12}\) = 0
\(\Rightarrow\left(x+1\right).\left(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\right)\)
Vì \(\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}\ne0\) nên x + 1 =0
=> x = -1
c) Xem lại đề
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bài 1:thực hiện phép tính : a)frac{3}{7}+frac{5}{13}+frac{4}{13}b)left(frac{3}{8}+frac{-3}{4}+frac{7}{12}right):frac{5}{6}+frac{1}{2}c)frac{2}{5}.frac{1}{3}-frac{2}{15}:frac{1}{5}+frac{3}{5}.frac{1}{3}d)left(4-frac{5}{12}right):2+frac{5}{24}e)frac{7}{19}.frac{8}{11}+frac{3}{11}.frac{7}{19}+frac{-1}{19}f)frac{9}{27}+frac{8}{24}+frac{18}{27}-frac{-16}{24}+frac{2}{3}g)frac{-5}{21}+frac{-2}{21}+frac{8}{24}h)frac{-5}{9}+frac{8}{15}+frac{-2}{11}+frac{4}{-9}+frac{7}{15}i)frac{7}{25}.frac{39}{-14}.frac{...
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bài 1:thực hiện phép tính :
a)\(\frac{3}{7}+\frac{5}{13}+\frac{4}{13}\)
b)\(\left(\frac{3}{8}+\frac{-3}{4}+\frac{7}{12}\right):\frac{5}{6}+\frac{1}{2}\)
c)\(\frac{2}{5}.\frac{1}{3}-\frac{2}{15}:\frac{1}{5}+\frac{3}{5}.\frac{1}{3}\)
d)\(\left(4-\frac{5}{12}\right):2+\frac{5}{24}\)
e)\(\frac{7}{19}.\frac{8}{11}+\frac{3}{11}.\frac{7}{19}+\frac{-1}{19}\)
f)\(\frac{9}{27}+\frac{8}{24}+\frac{18}{27}-\frac{-16}{24}+\frac{2}{3}\)
g)\(\frac{-5}{21}+\frac{-2}{21}+\frac{8}{24}\)
h)\(\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{4}{-9}+\frac{7}{15}\)
i)\(\frac{7}{25}.\frac{39}{-14}.\frac{50}{78}\)
m.n làm nhanh cho mình nhé, chiều mình phải nộp rồi! cảm ơn m.n!
Tính:
a) \(\frac{2}{7}:\frac{13}{24}+\frac{2}{7}:\frac{5}{24}\)
b)\(\frac{15}{24}-\frac{3}{8}-\frac{1}{6}\)
c)\(\frac{7}{13}.\frac{5}{14}.\frac{39}{15}\)
a)<=>2/7:(13/24+5/24) b)<=>15/24-9/24-4/24 c)=7.5.39/13.14.15
<=>2/7:3/4 <=>2/24=1/12 =3/2.3
<=>2/7x4/3 =1/2
<=>8/21
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a. \(\frac{2}{7}:\frac{13}{24}+\frac{2}{7}:\frac{5}{24}\)
= \(\frac{2}{7}:\left(\frac{13}{24}+\frac{5}{24}\right)\)
= \(\frac{2}{7}:\frac{3}{4}\)
= \(\frac{8}{21}\)
b. \(\frac{15}{24}-\frac{3}{8}-\frac{1}{6}\)
= \(\frac{15}{24}-\frac{9}{24}-\frac{4}{24}\)
= \(\frac{2}{24}=\frac{1}{12}\)
c. \(\frac{7}{13}.\frac{5}{14}.\frac{39}{15}\)
= \(\frac{7.5.3.13}{13.2.7.3.5}\)
= \(\frac{1}{2}\)
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\(a,\frac{2}{7}:\frac{13}{24}+\frac{2}{7}:\frac{5}{24}\)
\(=\frac{2}{7}\cdot\frac{24}{13}+\frac{2}{7}\cdot\frac{24}{5}\)
\(=\frac{2}{7}\left(\frac{24}{13}+\frac{24}{5}\right)\)
\(=\frac{2}{7}\cdot\frac{432}{65}\)
\(=\frac{864}{455}\)
\(b,\frac{15}{24}-\frac{3}{8}-\frac{1}{6}\)
\(=\frac{1}{4}-\frac{1}{6}\)
\(=\frac{1}{12}\)
\(c,\frac{7}{13}\cdot\frac{5}{24}\cdot\frac{39}{15}\)
\(=\frac{7\cdot5\cdot39}{13\cdot24\cdot15}\)
\(=\frac{7}{24}\)
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Đúng ghi Đ, sai ghi S vào chỗ chấm:a) frac{3}{7}+frac{1}{3}xfrac{3}{7}frac{16}{21}xfrac{3}{7}frac{16}{49} ......b)frac{5}{9}+frac{1}{5}xfrac{5}{6}frac{5}{9}xfrac{1}{6}frac{13}{18} ......c)frac{13}{9}-frac{7}{9}:frac{2}{3}frac{6}{9}:frac{3}{2}1 ......d)frac{11}{8}-frac{5}{8}:frac{3}{4}frac{11}{8}-frac{5}{8}xfrac{4}{3}frac{11}{8}-frac{10}{24}frac{23}{24} ......
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Đúng ghi Đ, sai ghi S vào chỗ chấm:
a) \(\frac{3}{7}+\frac{1}{3}x\frac{3}{7}=\frac{16}{21}x\frac{3}{7}=\frac{16}{49}\) ......
b)\(\frac{5}{9}+\frac{1}{5}x\frac{5}{6}=\frac{5}{9}x\frac{1}{6}=\frac{13}{18}\) ......
c)\(\frac{13}{9}-\frac{7}{9}:\frac{2}{3}=\frac{6}{9}:\frac{3}{2}=1\) ......
d)\(\frac{11}{8}-\frac{5}{8}:\frac{3}{4}=\frac{11}{8}-\frac{5}{8}x\frac{4}{3}=\frac{11}{8}-\frac{10}{24}=\frac{23}{24}\) ......
Mai Hồng Ngọc? Vũ Thị Thu Hằng? Ai đúng dzậy -_-*
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Tính giá trị biểu thức1.Afrac{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.Bleft(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.Cfrac{5}{18}+frac{8}{19}-frac{7}{21}+left(frac{-10}{36}+frac{11}{19}+frac{1}{3}right)-frac{5}{8}4.Dfrac{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.Efrac{1}{13}+left(frac{-5}{18}-frac...
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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
bài 1: cho x, y thuộc Q. cmr:
|x + y| =< |x| + |y|
bài 2: tính:
\(A=\frac{\left(13\frac{1}{4}-2\frac{5}{27}-10\frac{5}{6}\right).230\frac{1}{25}+46\frac{3}{4}}{\left(1\frac{3}{7}+\frac{10}{3}\right):\left(12\frac{1}{3}-14\frac{2}{7}\right)}\)
bài 3: cho a + b + c = a^2 + b^2 + c^2 = 1 và x : y : z = a : b : c.
cmr: (x + y + z)^2 = x^2 + y^2 + z^2
1
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Ez lắm =)
Bài 1:
Với mọi gt \(x,y\in Q\) ta luôn có:
\(x\le\left|x\right|\) và \(-x\le\left|x\right|\)
\(y\le\left|y\right|\) và \(-y\le\left|y\right|\Rightarrow x+y\le\left|x\right|+\left|y\right|\) và \(-x-y\le\left|x\right|+\left|y\right|\)
Hay: \(x+y\ge-\left(\left|x\right|+\left|y\right|\right)\)
Do đó: \(-\left(\left|x\right|+\left|y\right|\right)\le x+y\le\left|x\right|+\left|y\right|\)
Vậy: \(\left|x+y\right|\le\left|x\right|+\left|y\right|\)
Dấu "=" xảy ra khi: \(xy\ge0\)
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Bài 3:
Ta có: \(\frac{x}{a}=\frac{y}{b}=\frac{z}{c}=\frac{x+y+z}{a+b+c}=x+y+z\) (vì a + b + c = 1)
Do đó: \(\left(x+y+z\right)^2=\frac{x^2}{a^2}=\frac{y^2}{b^2}=\frac{z^2}{c^2}=\frac{x^2+y^2+z^2}{a^2+b^2+c^2}=x^2+y^2+z^2\) (vì a2 + b2 + c2 = 1)
Vậy: (x + y + z)2 = x2 + y2 + z2
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