a: \(2x+\frac12x=1\frac14\)

=>\(\frac52x=\frac54\)

=>\(x=\frac54:\frac52=\frac54\cdot\frac25=\frac24=\frac12\)

b: \(\frac12x+\frac13=2x-\frac15\)

=>\(\frac12x-2x=-\frac15-\frac13\)

=>\(-\frac32x=-\frac{8}{15}\)

=>\(x=\frac{8}{15}:\frac32=\frac{8}{15}\cdot\frac23=\frac{16}{45}\)

c: \(\left(\frac12x+\frac13\right)-\left(\frac14-\frac15x\right)=-1\frac15\)

=>\(\frac12x+\frac13-\frac14+\frac15x=-\frac65\)

=>\(\frac{7}{10}x=-\frac65-\frac13+\frac14=\frac{-72}{60}-\frac{20}{60}+\frac{15}{60}=\frac{-77}{60}\)

=>\(x=-\frac{11}{6}\)

d: \(\left(\frac12-\frac13x\right)+\left(-\frac14x+\frac12\right)-\left(1\frac13+\frac{-1}{4}x\right)=5\)

=>\(-\frac13x+\frac12-\frac14x+\frac12-\frac43+\frac14x=5\)

=>\(-\frac13x-\frac13=5\)

=>\(-\frac13x=5+\frac13=\frac{16}{3}\)

=>x=-16

e: \(\left(\frac12\left|x\right|+\frac13\right)\left(\frac14-\frac13x\right)=0\)

=>\(\frac14-\frac13x=0\)

=>\(\frac13x=\frac14\)

=>\(x=\frac14:\frac13=\frac34\)

f: \(\frac12x^2-\frac13x=0\)

=>\(x\left(\frac12x-\frac13\right)=0\)

=>\(\left[\begin{array}{l}x=0\\ \frac12x-\frac13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ \frac12x=\frac13\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac23\end{array}\right.\)

g: \(\frac23x^3+\frac14x=0\)

=>\(x\left(\frac23x^2+\frac14\right)=0\)

=>x=0

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)\(\dfrac{-5}{9}\).(\(\dfrac{3}{10}-\dfrac{1}{5}\))=\(\dfrac{-5}{9}\).(\(\dfrac{3}{10}-\dfrac{2}{10}\))=\(\dfrac{-5}{9}\).\(\dfrac{1}{10}\)=\(\dfrac{-1}{18}\)

Có đúng ko ạ

Bài 2:

\(a,\Rightarrow\left|\dfrac{3}{4}+x\right|=1\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}+x=1\\\dfrac{3}{4}+x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}\\x=-\dfrac{7}{4}\end{matrix}\right.\\ b,\Leftrightarrow x+\dfrac{2}{5}=\dfrac{4}{9}:\dfrac{4}{9}=1\Leftrightarrow x=\dfrac{3}{5}\)

b: \(\dfrac{4}{9}:\left(x+\dfrac{2}{5}\right)=\dfrac{4}{9}\)

\(\Leftrightarrow x+\dfrac{2}{5}=1\)

hay \(x=\dfrac{3}{5}\)

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) \(\Leftrightarrow\sqrt{2x}-4\sqrt{2x}+16\sqrt{2x}=52\left(đk:x\ge0\right)\)

\(\Leftrightarrow13\sqrt{2x}=52\Leftrightarrow\sqrt{2x}=4\Leftrightarrow2x=16\Leftrightarrow x=8\left(tm\right)\)

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

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:

\(sinG=\dfrac{HT}{GT}\Rightarrow sin35^0=\dfrac{y}{16}\Rightarrow y\approx9,18\)

\(cosG=\dfrac{GH}{GT}\Rightarrow cos35^0=\dfrac{x}{16}\Rightarrow x\approx10,11\)

Hình 1:

\(x=12\cdot\tan42^0\simeq10.8\left(cm\right)\)

\(y=\sqrt{10.8^2+12^2}\simeq16,14\left(cm\right)\)

a: Ta có: \(\sin^2a+cos^2a=1\)

=>\(cos^2a=1-0,6^2=0,64=0,8^2\)

=>cosa=0,8

\(\tan a=\frac{\sin a}{cosa}=\frac{0.6}{0.8}=\frac34\)

\(\cot a=\frac{1}{\tan a}=1:\frac34=\frac43\)

\(\sin\left(90^0-a\right)=cosa=0,8\)

\(cos\left(90^0-a\right)=\sin a=0,6\)

b: \(\sin^2a+cos^2a=1\)

=>\(sin^2a=1-\left(\frac{1}{\sqrt5}\right)^2=1-\frac15=\frac45\)

=>\(\sin a=\frac{2}{\sqrt5}\)

tan a=\(\frac{\sin a}{cosa}=\frac{2}{\sqrt5}:\frac{1}{\sqrt5}=2\)

cot a=1/tana=1/2

\(\tan\left(90^0-a\right)=\cot a=\frac12\)

\(\cot\left(90^0-a\right)=\tan a=2\)

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;

}

\(F_A=P-P'=12-9=3\left(N\right)\)

\(=>V=\dfrac{F_A}{d}=\dfrac{3}{10000}=3.\cdot10^{-4}m^3\)

B có thể làm đủ cả 3 câu o ạ? Tks

a: Để (d) cắt (d') thì \(\frac{1}{m}<>-m\)

=>\(-m^2<>1\)

=>\(m^2<>-1\) (luôn đúng)

=>(d) luôn cắt (d')