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27 tháng 10 2021
a) (1,75 : \(\dfrac{7}{2}\)).\(\dfrac{8}{5}\)=(\(\dfrac{7}{4}\) : \(\dfrac{7}{2}\)).\(\dfrac{8}{5}\)=(\(\dfrac{7}{4}\).\(\dfrac{2}{7}\)).\(\dfrac{8}{5}\)=\(\dfrac{1}{2}\).\(\dfrac{8}{5}\)=\(\dfrac{4}{5}\)
b) \(\dfrac{7}{2}\).\(4\dfrac{5}{3}\)-\(2\dfrac{5}{3}\).\(\dfrac{7}{2}\)=(\(4\dfrac{5}{3}\)-\(2\dfrac{5}{3}\)).\(\dfrac{7}{2}\)=2.\(\dfrac{7}{2}\)=7
c)\(\dfrac{-5}{9}\).(\(\dfrac{3}{10}-\dfrac{1}{5}\))=\(\dfrac{-5}{9}\).(\(\dfrac{3}{10}-\dfrac{2}{10}\))=\(\dfrac{-5}{9}\).\(\dfrac{1}{10}\)=\(\dfrac{-1}{18}\)
27 tháng 10 2021
Bài 2:
\(a,\Rightarrow\left|\dfrac{3}{4}+x\right|=1\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}+x=1\\\dfrac{3}{4}+x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{7}{4}\end{matrix}\right.\\ b,\Leftrightarrow x+\dfrac{2}{5}=\dfrac{4}{9}:\dfrac{4}{9}=1\Leftrightarrow x=\dfrac{3}{5}\)
27 tháng 10 2021
b: \(\dfrac{4}{9}:\left(x+\dfrac{2}{5}\right)=\dfrac{4}{9}\)
\(\Leftrightarrow x+\dfrac{2}{5}=1\)
hay \(x=\dfrac{3}{5}\)
YN
2
22 tháng 9 2021
a) \(\Leftrightarrow x^2=\sqrt{4}\)
\(\Leftrightarrow x^2=2\Leftrightarrow x=\pm2\)
b) \(\Leftrightarrow\sqrt{\left(\dfrac{1}{2}x+1\right)^2}=9\)
\(\Leftrightarrow\left|\dfrac{1}{2}x+1\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+1=9\\\dfrac{1}{2}x+1=-9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=-16\end{matrix}\right.\)
c) \(\Leftrightarrow\sqrt{2x}-4\sqrt{2x}+16\sqrt{2x}=52\left(đk:x\ge0\right)\)
\(\Leftrightarrow13\sqrt{2x}=52\Leftrightarrow\sqrt{2x}=4\Leftrightarrow2x=16\Leftrightarrow x=8\left(tm\right)\)
22 tháng 9 2021
f: Ta có: \(\sqrt{\dfrac{50-25x}{4}}-8\sqrt{2-x}+\sqrt{18-9x}=-10\)
\(\Leftrightarrow\sqrt{2-x}\cdot\dfrac{5}{2}-8\sqrt{2-x}+3\sqrt{2-x}=-10\)
\(\Leftrightarrow\sqrt{2-x}=4\)
\(\Leftrightarrow2-x=16\)
hay x=-14
6 tháng 10 2021
Hình 1:
Áp dụng tslg:
\(cosK=\dfrac{IK}{MK}\)\(\Rightarrow cos42^0=\dfrac{12}{y}\Rightarrow y\approx16,15\)
\(tanK=\dfrac{IM}{IK}\Rightarrow tan42^0=\dfrac{x}{12}\Rightarrow x\approx10,8\)
Hình 2:
\(sinG=\dfrac{HT}{GT}\Rightarrow sin35^0=\dfrac{y}{16}\Rightarrow y\approx9,18\)
\(cosG=\dfrac{GH}{GT}\Rightarrow cos35^0=\dfrac{x}{16}\Rightarrow x\approx10,11\)
6 tháng 10 2021
Hình 1:
\(x=12\cdot\tan42^0\simeq10.8\left(cm\right)\)
\(y=\sqrt{10.8^2+12^2}\simeq16,14\left(cm\right)\)
V
0
25 tháng 10 2021
Câu 31:
#include <bits/stdc++.h>
using namespace std;
long long n,i,x,dem;
int main()
{
cin>>n;
dem=0;
for (i=1; i<=n; i++)
{
cin>>x;
if (30%x==0) dem++;
}
cout<<dem;
return 0;
}
V
0
\(a,\dfrac{6.12-6.7}{60}=\dfrac{6\left(12-7\right)}{60}=\dfrac{6.5}{60}=\dfrac{30}{60}=\dfrac{1}{2}\\ b,\dfrac{35.18-35}{\left(-34\right).7}=\dfrac{35\left(18-1\right)}{\left(-34\right).7}=\dfrac{35.17}{\left(-34\right).7}=\dfrac{595}{-238}=\dfrac{-5}{2}\)
\(c,\dfrac{42-11.42}{21.\left(-15\right)}=\dfrac{42\left(1-11\right)}{21.\left(-15\right)}=\dfrac{2.21.\left(-10\right)}{21.\left(-15\right)}=\dfrac{-20}{-15}=\dfrac{4}{3}\)