giúp bài 3, câu c,f,g,l,k bài 2
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Những câu hỏi liên quan
S
16 tháng 12 2022
a)=-50
b)=-33
c)=13
d)=46
e)=-33
f)=-54
g)=29
h)=-11
i)=-120
k)=-196
l)=-108
m)=100
HP
16 tháng 12 2022
a)=-50
b)=-33
c)=13
d)=46
e)=-33
f)=-54
g)=29
h)=-11
i)=-120
k)=-196
l)=-108
m)=100
NQ
7
HM
3
KK
7 tháng 9 2021
a. \(\sqrt{12^2}\)
= 12
b. \(\sqrt{\left(-7\right)^2}\)
= 7
c. \(\sqrt{\left(2-\sqrt{5}\right)^2}\)
= 2 - \(\sqrt{5}\)
HP
23 tháng 9 2021
j, ĐK: \(x\ne\dfrac{\pi}{6}+\dfrac{k\pi}{2}\)
\(tan\left(\dfrac{\pi}{3}+x\right)-tan\left(\dfrac{\pi}{6}+2x\right)=0\)
\(\Leftrightarrow tan\left(\dfrac{\pi}{3}+x\right)=tan\left(\dfrac{\pi}{6}+2x\right)\)
\(\Leftrightarrow\dfrac{\pi}{3}+x=\dfrac{\pi}{6}+2x+k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{6}+k\pi\left(l\right)\)
\(\Rightarrow\) vô nghiệm.
4 tháng 4 2020
Mik biết 1 câu mấy
1. \(f\left(1\right)=a.1^2+b.1+6\)
\(=a+b+6=3\)
\(=a+6=6-3\)
\(=a+b=3\)
Để đa thức f(x) có bậc là 1 thì a phải là 0
Vậy a=0 và b= -3
LT
2
18 tháng 9 2021
c) \(=\left(5a-\dfrac{1}{3}\right)^2\)
d) \(=\left(y-\dfrac{1}{3}\right)^2\)
e) \(=\left(2x-y+1\right)^2\)
f) \(=\left(2x-4y\right)^2+2\left(2x-4y\right)+1=\left(2x-4y+1\right)^2\)
g) \(=\left(2xy^2-3\right)^2\)
18 tháng 9 2021
\(c,=\left(5a-\dfrac{1}{3}\right)^2\\ d,=\left(y^4-\dfrac{1}{3}\right)^2\\ e,=\left(2x-y+1\right)^2\\ f,=\left(2x-4y\right)^2+4\left(x-2y\right)+1=\left(2x-4y+1\right)\\ g,=\left(2xy^2-3\right)^2\)
Bài 2:
c: \(\left(\dfrac{3}{2}+x\right):1\dfrac{2}{5}=\dfrac{1}{2}\cdot\dfrac{3}{5}+0,2\)
=>\(\left(x+\dfrac{3}{2}\right):\dfrac{7}{5}=\dfrac{3}{10}+\dfrac{1}{5}=\dfrac{5}{10}=\dfrac{1}{2}\)
=>\(x+\dfrac{3}{2}=\dfrac{1}{2}\cdot\dfrac{7}{5}=\dfrac{7}{10}\)
=>\(x=\dfrac{7}{10}-\dfrac{3}{2}=\dfrac{7}{10}-\dfrac{15}{10}=-\dfrac{8}{10}=-\dfrac{4}{5}\)
f: \(-\dfrac{7}{5}-\left(\dfrac{2}{3}+x\right)=\dfrac{3}{10}\)
=>\(-\dfrac{7}{5}-\dfrac{2}{3}-x=\dfrac{3}{10}\)
=>\(x=-\dfrac{7}{5}-\dfrac{2}{3}-\dfrac{3}{10}=\dfrac{-42-20-9}{30}=\dfrac{-71}{30}\)
g: \(\left|x-\dfrac{1}{6}\right|-\dfrac{7}{12}=\dfrac{1}{4}\)
=>\(\left|x-\dfrac{1}{6}\right|=\dfrac{1}{4}+\dfrac{7}{12}=\dfrac{10}{12}=\dfrac{5}{6}\)
=>\(\left[{}\begin{matrix}x-\dfrac{1}{6}=\dfrac{5}{6}\\x-\dfrac{1}{6}=-\dfrac{5}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{6}=1\\x=-\dfrac{5}{6}+1=-\dfrac{4}{6}=-\dfrac{2}{3}\end{matrix}\right.\)
l: \(x:\left(\dfrac{1}{5}-\dfrac{7}{10}\right)=-2+\left(-1\dfrac{2}{3}\right)\)
=>\(x:\left(\dfrac{2}{10}-\dfrac{7}{10}\right)=-2+\dfrac{-5}{3}=\dfrac{-11}{3}\)
=>\(x:\dfrac{-1}{2}=-\dfrac{11}{3}\)
=>\(x=\dfrac{11}{3}\cdot\dfrac{1}{2}=\dfrac{11}{6}\)
k: \(x:\left(\dfrac{1}{7}-\dfrac{3}{14}\right)=-3+\left(-1\dfrac{2}{3}\right)\)
=>\(x:\left(\dfrac{2}{14}-\dfrac{3}{14}\right)=-3-\dfrac{5}{3}=\dfrac{-14}{3}\)
=>\(x:\dfrac{-1}{14}=\dfrac{-14}{3}\)
=>\(x=\dfrac{14}{3}\cdot\dfrac{1}{14}=\dfrac{1}{3}\)
Bài 3:
a: \(A=\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}\)
\(=\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}\)
\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{7}-\dfrac{1}{8}\)
\(=\dfrac{1}{2}-\dfrac{1}{8}=\dfrac{3}{8}\)
b: \(B=\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\)
\(=\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\)
\(=\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=\dfrac{1}{4}-\dfrac{1}{10}=\dfrac{5-2}{20}=\dfrac{3}{20}\)