A) x = 5

B) x = 18

C) thì ko bít

a. \(\dfrac{x}{5}=\dfrac{25}{75}\)

\(x:5=25:75=\dfrac{1}{3}\)

\(x=\dfrac{1}{3}\text{×}5=\dfrac{5}{3}\)

b. \(\dfrac{16}{x}=\dfrac{8}{9}\)

\(16:x=8:9\)

\(x=16:8:9=\dfrac{2}{9}\)

c. \(\dfrac{20}{35}=\dfrac{x}{14}\)

\(x:14=20:35=\dfrac{4}{7}\)

\(x=\dfrac{4}{7}\text{×}14=8\)

Lời giải:
a.

$x=\frac{25}{75}\times 5=\frac{25\times 5}{75}=\frac{25\times 5}{25\times 3}=\frac{5}{3}$

b.

$x=16: \frac{8}{9}=16\times \frac{9}{8}=\frac{8\times 2\times 9}{8}=18$

c.

$x=\frac{20}{35}\times 14=\frac{20\times 14}{35}=\frac{4\times 5\times 7\times 2}{5\times 7}=4\times 2=8$

bài 1

\(\widehat{B}=90-\widehat{C}=90-30=60\)

\(sinC=\dfrac{AB}{BC}\Rightarrow BC=\dfrac{AB}{sinC}=\dfrac{30}{sin30}=60\)

áp dụng pytago vào \(\Delta ABC\)

\(AC=\sqrt{BC^2-AB^2}\)=\(\sqrt{60^2-30^2}\)=\(30\sqrt{3}\)=51,96

bài 2

\(\widehat{B}=90-\widehat{C}=90-30=60\)

\(sinC=\dfrac{AB}{BC}\Rightarrow AB=sinC.BC=sin30.5=2,5\)

áp 

áp dụng pytago vào \(\Delta ABC\)

\(AC=\sqrt{BC^2-AB^2}=\sqrt{5^2-2,5^2}\)=4,33

bài 3

\(\widehat{E}=90-\widehat{F}=90-47=43\)

\(sinF=\dfrac{ED}{EF}\Rightarrow EF=\dfrac{ED}{sinF}=\dfrac{9}{sin47}=12,31\)

áp dụng pytago vào \(\Delta DEF\)

\(DF=\sqrt{EF^2-ED^2}=\sqrt{12,31^2-9^2}\)=8,4

bài 4

áp dụng pytago vào \(\Delta ABC\)

\(AB=\sqrt{BC^2-AC^2}=\sqrt{32^2-27^2}=17,18\)

\(sinB=\dfrac{AC}{BC}=\dfrac{27}{32}\Rightarrow\widehat{B}=57\)

\(\widehat{C}=90-\widehat{B}=90-57=33\)

#include <bits/stdc++.h>

using namespace std;

int main()

{

long long a,c,tam;

cin>>a>>c;

cout<<"a truoc khi doi la:"<<a<<endl;

cout<<"c truoc khi doi la:"<<c<<endl;

tam=a;

a=c;

c=tam;

cout<<"a sau khi doi la:"<<a<<endl;

cout<<"c sau khi doi la:"<<c<<endl;

return 0;

}

nhìu vậy mik trl một ít thui có đc ko

câu 6 : B cÂU 8 : c

6 b 7 d 8 mik ko bt 9 d 10 mik ko bt 11 d 12 b 13 d 14 d 15 b 16 b 17 b 18 b 19 a 20 b

nhưng cậu kết bạn với tui thì tui giải

Ta có:

\(\widehat{A}=180^o-60^o-45^o=75^o\)(tổng 3 góc trong tam giác)

Kẻ đường cao AH

\(sin60=\dfrac{AH}{AB}\)

\(\Rightarrow AH=sin60.4=2\sqrt{3}\)

\(sin45=\dfrac{AH}{AC}\)

\(\Rightarrow AC=\dfrac{2\sqrt{3}}{sin45}=2\sqrt{6}\)

\(BC=\sqrt{4^2-\left(2\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{6}\right)^2-\left(2\sqrt{3}\right)^2}=2+2\sqrt{3}\)

`(2x - 1)^2 = 36`

`(2x - 1)^2 = 6^2` hoặc `(2x - 1)^2 = (-6)^2`

`2x - 1 = 6` hoặc `2x - 1 = -6`

`2x = 6 + 1` hoặc `2x = -6 + 1`

`2x = 7` hoặc `2x = -5`

`x = 7 : 2` hoặc `x = -5 : 2`

`x = 3,5` hoặc `x = -2,5`

\(\left(2x-1\right)^2=36\)

\(\Rightarrow\left(2x-1\right)^2=6^2\)

\(\Rightarrow\left[{}\begin{matrix}2x-1=6\\2x-1=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=7\\2x=-5\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

\(P=x^2-4x+2x-8+9,5=x^2-2x+1-9+9,5=\)

\(=\left(x-1\right)^2+0,5>0\forall x\)