a,tính.7 phần 9+9 phần 10=
b,tìm y y+2 phần 7=1 1 phần 2
giúp mk vs mk đang cần gấp
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\(a.\frac{2}{3}+\frac{1}{5}\cdot\frac{10}{7}\)
\(=\frac{2}{3}+\frac{1\cdot2}{1\cdot7}\)
\(=\frac{2}{3}+\frac{2}{7}=\frac{2\cdot7}{21}+\frac{2\cdot3}{21}=\frac{14}{21}+\frac{6}{21}=\frac{20}{21}\)
\(b.\frac{2}{7}\cdot\frac{4}{7}+\frac{2}{7}\cdot\frac{3}{7}\)
\(=\frac{2}{7}\cdot\left(\frac{4}{7}+\frac{3}{7}\right)\)
\(=\frac{2}{7}\cdot\frac{7}{7}=\frac{2}{7}\cdot1=\frac{2}{7}\)
\(c.\left[-\frac{1}{4}+\frac{3}{10}\right]:\left(-\frac{3}{5}\right)-\frac{7}{6}\)
\(=\left[-\frac{5}{20}+\frac{6}{20}\right]:\left(-\frac{3}{5}\right)-\frac{7}{6}\)
\(=\frac{1}{20}:\left(-\frac{3}{5}\right)-\frac{7}{6}\)
\(=\frac{1}{20}\cdot\left(-\frac{5}{3}\right)-\frac{7}{6}\)
\(=\frac{1\cdot\left(-1\right)}{4\cdot3}-\frac{7}{6}\)
\(=\left(-\frac{1}{12}\right)-\frac{7}{6}=\left(-\frac{1}{12}\right)-\frac{14}{12}=-\frac{15}{12}\)
bài 2 :a) \(2x-2\frac{2}{7}=2\frac{5}{7}\)
\(2x=2\frac{5}{7}-2\frac{2}{7}\)
\(2x=\left(2-2\right)+\left(\frac{5}{7}-\frac{2}{7}\right)\)
\(2x=0+\frac{3}{7}\)
\(2x=\frac{3}{7}\)
\(x=\frac{3}{7}:2=\frac{3}{7}\cdot\frac{1}{2}=\frac{3}{14}\)
\(b.\frac{17}{13x}=\frac{4}{39}\)
\(\Rightarrow13x\cdot4=17\cdot39\)
\(\Rightarrow13x\cdot4=663\)
\(\Rightarrow13x=663:4\)
\(\Rightarrow13x=165,75\)
\(\Rightarrow x=165,75:13\)
\(\Rightarrow x=12,75\)
\(****nha!!!!!!!!!!!!\)
I don't now
or no I don't
..................
sorry
a) Ta có: \(M=\left(\dfrac{1}{2}x^2y\right)\cdot\left(\dfrac{2}{3}xy\right)^2\)
\(=\dfrac{1}{2}x^2y\cdot\dfrac{4}{9}x^2y^2\)
\(=\dfrac{2}{9}x^4y^3\)
b) Hệ số là \(\dfrac{2}{9}\)
Phần biến là \(x^4;y^3\)
c) Bậc là 7
d) Thay x=-1 và y=2 vào M, ta được:
\(M=\dfrac{2}{9}\cdot\left(-1\right)^4\cdot2^3=\dfrac{2}{9}\cdot8=\dfrac{16}{9}\)
a: =>-2x=90/91
hay x=-45/91
b: =>2x=-7
hay x=-7/2
c: ->-3x=-12
hay x=4
Bài 1:
a: =>2x-9=10/91
=>2x=829/91
hay x=829/182
b: =>2x=-7
hay x=-7/2
c: =>-3x=-12
hay x=4
(1/2)^m = 1/32
mà 1/32 = (1/2)^5 nên m = 5
343/125= (7/5)^n
mà 343/125 = (7/5)^3 nên n=3
a: \(M=\left(\dfrac{-3}{7}x^3y\right)\cdot\dfrac{7xy^3}{12}-x^2y^2\cdot\left(-\dfrac{3}{4}x^2y^2\right)\)
\(=\dfrac{-1}{4}x^4y^4+\dfrac{3}{4}x^4y^4\)
\(=\dfrac{1}{2}x^4y^4\)
b: Hệ số là 1/2
Biến là \(x^4;y^4\)
bậc là 4+4=8
c: Thay x=-1 và y=-2 vào M, ta được:
\(M=\dfrac{1}{2}\left(-1\right)^4\cdot\left(-2\right)^4=\dfrac{1}{2}\cdot16=8\)
a) 7/10 - y × 3/4 = 1/5
y × 3/4 = 7/10 - 1/5
y × 3/4 = 1/2
y = 1/2 : 3/4
y = 2/3. Vậy y = 2/3
b) 5/6 : ( y + 7/9 ) = 3/4
y + 7/9 = 5/6 : 3/4
y + 7/9 = 10/9
y. = 10/9 - 7/9
y = 1/3. Vậy y = 1/3
\(\frac{7}{10}\)- y * \(\frac{3}{4}\)= \(\frac{1}{5}\)
y * \(\frac{3}{4}\)= \(\frac{7}{10}\)- \(\frac{1}{5}\)
y * \(\frac{3}{4}\)= \(\frac{1}{2}\)
y = \(\frac{1}{2}\): \(\frac{3}{4}\)
y = \(\frac{2}{3}\)
Vậy y = \(\frac{2}{3}\)
\(\frac{5}{6}\): ( y + \(\frac{7}{9}\)) = \(\frac{3}{4}\)
y + \(\frac{7}{9}\) = \(\frac{5}{6}\): \(\frac{3}{4}\)
y + \(\frac{7}{9}\) = \(\frac{10}{9}\)
y = \(\frac{10}{9}\)- \(\frac{7}{9}\)
y = \(\frac{3}{9}\)
Vậy y = \(\frac{3}{9}\)
a)7/9+7/10=133/90
b)y+2/7 = 11/2
y = 11/2 - 2/7
y=73/14