b: \(cos14=sin76;cos37=sin53\)

Vì 47<48<53<76 nên \(sin47< sin48< sin53< sin76\)

=>\(sin47< sin48< cos37< cos14\)

c: \(cot25=tan65;cot38=tan52\)

Vì 52<62<65<73

nên \(tan52< tan62< tan65< tan73\)

=>\(cot38< tan63< cot25< tan73\)

a: Vì 75>45

nên \(tan75>1\)

Vì 40<45

nên \(cot40>1\)

\(cot40=tan50;tan75=tan75\)

mà \(tan50< tan75\)

nên \(1< cot40< tan75\left(1\right)\)

\(cos56=sin34;sin50=sin50\)

mà 34<50

nên \(sin34< sin50< 1\)

=>\(cos56< sin50< 1\left(2\right)\)

Từ (1),(2) suy ra \(cos56< sin50< cot40< tan75\)

Gọi tốc độ lượt đi của ô tô là: `x` (km/h)

ĐK: `x > 0` 

Thời gian đi từ A đến B của ô tô là: \(\dfrac{160}{x}\left(h\right)\)

Tốc độ lượt về của ô tô là: \(\left(100\%-20\%\right)x=80\%x=0,8x\left(km/h\right)\)

Thời gian về của ô tô là: \(\dfrac{160}{0,8x}=\dfrac{200}{x}\left(h\right)\)

Tổng thời gian đi và về của ô tô là 7 giờ 12 phút hay `36/5` giờ nên ta có pt:

\(\dfrac{160}{x}+\dfrac{200}{x}=\dfrac{36}{5}\)

\(\Leftrightarrow\dfrac{160+200}{x}=\dfrac{36}{5}\)

\(\Leftrightarrow\dfrac{360}{x}=\dfrac{36}{5}\)

\(\Leftrightarrow x=360:\dfrac{36}{5}\)

\(\Leftrightarrow x=50\left(tm\right)\)

Vậy: ... 

Xét ΔABC vuông tại A có \(\widehat{B}+\widehat{C}=90^0\)

nên \(sinB=cosC=\dfrac{4}{5}\)

\(sin^2B+cos^2B=1\)

=>\(cos^2B=1-\left(\dfrac{4}{5}\right)^2=\dfrac{9}{25}=\left(\dfrac{3}{5}\right)^2\)

=>\(cosB=\dfrac{3}{5}\)

\(tanB=\dfrac{sinB}{cosB}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\)

\(cotB=\dfrac{1}{tanB}=\dfrac{3}{4}\)

Vì tam giác ABC vuông tại A

Nên: \(\widehat{B}+\widehat{C}=90^o\\ \Rightarrow0^o< \widehat{C}< 90^o\)

\(\Rightarrow0< \sin C< 1\) 

Ta có: \(\sin^2C+\cos^2C=1\Rightarrow\sin^2C=1-\left(\dfrac{4}{5}\right)^2=\dfrac{9}{25}\\ \Rightarrow\sin C=\dfrac{3}{5}\)

Lại có: \(\tan C=\dfrac{\sin C}{\cos C}=\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}=\dfrac{3}{4}\\ \cot C=\dfrac{1}{\tan C}=\dfrac{4}{3}\)

\(5x^2-2x+1=\left(4x-1\right)\sqrt{x^2}+1\)

\(\Rightarrow5x^2-2x=\left(4x-1\right)x\)

\(\Rightarrow5x^2-2x=4x^2-x\)

\(\Rightarrow5x^2-4x^2-2x+x=0\)

\(\Rightarrow x^2-x=0\)

\(\Rightarrow x\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy x=0 hoặc x=1

Bài 17:

$a^2-2a(b+c)=b^2-2b(c+a)$
$\Leftrightarrow a^2-2ac=b^2-2bc$

$\Leftrightarrow (a^2-b^2)-(2ac-2bc)=0$

$\Leftrightarrow (a-b)(a+b-2c)=0$

$\Leftrightarrow a=b$ hoặc $a+b=2c$.

Nếu $a=b$. Thay vào đk $b^2-2b(c+a)=c^2-2c(a+b)$ thì:

$a^2-2a(c+a)=c^2-2c(a+a)$
$\Leftrightarrow -a^2-2ac=c^2-4ac$

$\Leftrightarrow a^2+c^2-2ac=0\Leftrightarrow (a-c)^2=0$
$\Leftrightarrow a=c$

Vậy $a=b=c\Rightarrow M=0$

Nếu $a+b=2c$

Khi đó ta có:

$a^2-2a(b+c)+b^2-2b(c+a)=2c^2-4c(a+b)$

$\Leftrightarrow a^2+b^2-4ab-2c(a+b)=2c^2-4c(a+b)$
$\Leftrightarrow (a+b)^2-6ab=2c^2-2c(a+b)=2c^2-2c.2c=-2c^2$

$\Leftrightarrow 4c^2-6ab=-2c^2$

$\Leftrightarrow 6ab=6c^2$

$\Leftrightarrow ab=c^2$

$\Leftrightarrow 4ab=4c^2=(2c)^2=(a+b)^2$

$\Leftrightarrow 4ab=a^2+b^2+2ab\Leftrightarrow (a-b)^2=0$

$\Leftrightarrow a=b$

Khi đó lại quay về TH1 và ta lại cm được $a=c$ nữa.

$\Rightarrow a=b=c\Rightarrow M=0$

Vậy $M=0$

Bài 18:

Đặt $\frac{a}{b-c}=x, \frac{b}{c-a}=y, \frac{c}{a-b}=z$.

Khi đó:

$xy+yz+xz=\frac{ab}{(b-c)(c-a)}+\frac{ac}{(b-c)(a-b)}+\frac{bc}{(c-a)(a-b)}=\frac{ab(a-b)+ac(a-c)+bc(b-c)}{(a-b)(b-c)(c-a)}=-1$

$N=(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})$

$=(x+y+z).\frac{xy+yz+xz}{xyz}=-\frac{x+y+z}{xyz}$

$=-[\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}]$

$=-[\frac{(b-c)(c-a)}{ab}+\frac{(c-a)(a-b)}{bc}+\frac{(b-c)(a-b)}{ac}]$

$=-\frac{c(b-c)(c-a)+a(c-a)(a-b)+b(b-c)(a-b)}{abc}$

$=\frac{a^3+b^3+c^3-ab(a+b)-bc(b+c)-ac(a+c)+3abc}{abc}$
$=\frac{a^3+b^3+c^3-ab(-c)-bc(-a)-ac(-b)-3abc}{abc}$

$=\frac{a^3+b^3+c^3+6abc}{abc}$

$=\frac{(a+b)^3-3ab(a+b)+c^3+6abc}{abc}=\frac{(-c)^3-3ab(-c)+c^3+6abc}{abc}$

$=\frac{-c^3+3abc+c^3+6abc}{abc}=\frac{9abc}{abc}=9$

\(a+b=1\)

=>a=1-b

\(\dfrac{a}{b^3-1}+\dfrac{b}{a^3-1}=\dfrac{1-b}{\left(b-1\right)\left(b^2+b+1\right)}+\dfrac{b}{\left(1-b\right)^3-1}\)

\(=\dfrac{-1}{b^2+b+1}+\dfrac{b}{\left(1-b-1\right)\left[\left(1-b\right)^2+1\left(1-b\right)+1\right]}\)

\(=\dfrac{-1}{b^2+b+1}+\dfrac{-1}{b^2-2b+1+1-b+1}\)

\(=\dfrac{-1}{b^2+b+1}+\dfrac{-1}{b^2-3b+3}=\dfrac{-b^2+3b-3-b^2-b-1}{\left(b^2+b+1\right)\left(b^2-3b+3\right)}\)

\(=\dfrac{-2b^2+2b-4}{\left(b^2+b+1\right)\left(b^2-3b+3\right)}\)

\(=\dfrac{-2b^2+2b-4}{b^4-3b^3+3b^2+b^3-3b^2+3b+b^2-3b+3}\)

\(=\dfrac{-2b^2+2b-4}{b^4-2b^3+b^2+3}\)(2)

\(\dfrac{2\left(ab-2\right)}{a^2b^2+3}\)

\(=\dfrac{2\left[b\left(1-b\right)-2\right]}{\left(1-b\right)^2b^2+3}=\dfrac{2\left[b-b^2-2\right]}{\left(b-b^2\right)^2+3}\)

\(=\dfrac{-2b^2+2b-4}{b^4-2b^3+b^2+3}\)(1)

Từ (1),(2) suy ra \(\dfrac{a}{b^3-1}+\dfrac{b}{a^3-1}=\dfrac{2\left(ab-2\right)}{a^2b^2+3}\)

\(a+b=1\)

=>a=1-b

\(\dfrac{a}{b^3-1}+\dfrac{b}{a^3-1}=\dfrac{1-b}{\left(b-1\right)\left(b^2+b+1\right)}+\dfrac{b}{\left(1-b\right)^3-1}\)

\(=\dfrac{-1}{b^2+b+1}+\dfrac{b}{\left(1-b-1\right)\left[\left(1-b\right)^2+1\left(1-b\right)+1\right]}\)

\(=\dfrac{-1}{b^2+b+1}+\dfrac{-1}{b^2-2b+1+1-b+1}\)

\(=\dfrac{-1}{b^2+b+1}+\dfrac{-1}{b^2-3b+3}=\dfrac{-b^2+3b-3-b^2-b-1}{\left(b^2+b+1\right)\left(b^2-3b+3\right)}\)

\(=\dfrac{-2b^2+2b-4}{\left(b^2+b+1\right)\left(b^2-3b+3\right)}\)

\(=\dfrac{-2b^2+2b-4}{b^4-3b^3+3b^2+b^3-3b^2+3b+b^2-3b+3}\)

\(=\dfrac{-2b^2+2b-4}{b^4-2b^3+b^2+3}\)(2)

\(\dfrac{2\left(ab-2\right)}{a^2b^2+3}\)

\(=\dfrac{2\left[b\left(1-b\right)-2\right]}{\left(1-b\right)^2b^2+3}=\dfrac{2\left[b-b^2-2\right]}{\left(b-b^2\right)^2+3}\)

\(=\dfrac{-2b^2+2b-4}{b^4-2b^3+b^2+3}\)(1)

Từ (1),(2) suy ra \(\dfrac{a}{b^3-1}+\dfrac{b}{a^3-1}=\dfrac{2\left(ab-2\right)}{a^2b^2+3}\)

\(a+b=1\)

=>a=1-b

\(\dfrac{a}{b^3-1}+\dfrac{b}{a^3-1}=\dfrac{1-b}{\left(b-1\right)\left(b^2+b+1\right)}+\dfrac{b}{\left(1-b\right)^3-1}\)

\(=\dfrac{-1}{b^2+b+1}+\dfrac{b}{\left(1-b-1\right)\left[\left(1-b\right)^2+1\left(1-b\right)+1\right]}\)

\(=\dfrac{-1}{b^2+b+1}+\dfrac{-1}{b^2-2b+1+1-b+1}\)

\(=\dfrac{-1}{b^2+b+1}+\dfrac{-1}{b^2-3b+3}=\dfrac{-b^2+3b-3-b^2-b-1}{\left(b^2+b+1\right)\left(b^2-3b+3\right)}\)

\(=\dfrac{-2b^2+2b-4}{\left(b^2+b+1\right)\left(b^2-3b+3\right)}\)

\(=\dfrac{-2b^2+2b-4}{b^4-3b^3+3b^2+b^3-3b^2+3b+b^2-3b+3}\)

\(=\dfrac{-2b^2+2b-4}{b^4-2b^3+b^2+3}\)(2)

\(\dfrac{2\left(ab-2\right)}{a^2b^2+3}\)

\(=\dfrac{2\left[b\left(1-b\right)-2\right]}{\left(1-b\right)^2b^2+3}=\dfrac{2\left[b-b^2-2\right]}{\left(b-b^2\right)^2+3}\)

\(=\dfrac{-2b^2+2b-4}{b^4-2b^3+b^2+3}\)(1)

Từ (1),(2) suy ra \(\dfrac{a}{b^3-1}+\dfrac{b}{a^3-1}=\dfrac{2\left(ab-2\right)}{a^2b^2+3}\)

\(a)\dfrac{x+5}{x-3}+2=\dfrac{2}{x-3}\left(x\ne3\right)\) 

`<=>` \(\dfrac{x+5}{x-3}+\dfrac{2\left(x-3\right)}{x-3}=\dfrac{2}{x-3}\)

`<=>` \(x+5+2\left(x-3\right)=2\)

`<=>` `x+5+2x-6=2`

`<=>x+2x=2-5+6`

`<=>3x=3`

`<=>x=1`

Vậy phương trình có nghiệm `x=1`

\(b)\dfrac{2}{x-2}+\dfrac{3}{x-3}=\dfrac{3x-20}{\left(x-2\right)\left(x-3\right)}\left(x\ne2;3\right)\)

`<=>` \(\dfrac{2\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{3\left(x-2\right)}{\left(x-3\right)\left(x-2\right)}=\dfrac{3x-20}{\left(x-2\right)\left(x-3\right)}\)

`<=>` \(2\left(x-3\right)+3\left(x-2\right)=3x-20\)

`<=>` `2x-6+3x-6=3x-20`

`<=>` `2x+3x-3x=-20+6+6`

`<=>` `2x=-8`

`<=>` `x=-4`

Vậy phương trình có nghiệm `x=-4`

\(c)\dfrac{3}{x-2}+\dfrac{2}{x+1}=\dfrac{2x+5}{\left(x-2\right)\left(x+1\right)}\left(x\ne2;-1\right)\)

`<=>` \(\dfrac{3\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}+\dfrac{2\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{2x+5}{\left(x-2\right)\left(x+1\right)}\)

`<=>` `3(x+1)+2(x-2)=2x+5`

`<=>` `3x+3+2x-4=2x+5`

`<=>` `3x+2x-2x=5-3+4`

`<=>` `3x=6`

`<=>` `x=2` (không thỏa mãn)

`<=>` Vô nghiệm

Vậy phương trình vô nghiệm.

a) \(\dfrac{x+5}{x-3}+2=\dfrac{2}{x-3}\left(x\ne3\right)\)

\(\Leftrightarrow\dfrac{x+5}{x-3}+\dfrac{2\left(x-3\right)}{x-3}=\dfrac{2}{x-3}\)

\(\Leftrightarrow x+5+2\left(x-3\right)=2\)

\(\Leftrightarrow x+5+2x-6=2\)

\(\Leftrightarrow3x-1=2\)

\(\Leftrightarrow3x=3\)

\(\Leftrightarrow x=1\left(tm\right)\)

Vậy: ... 

b) \(\dfrac{2}{x-2}+\dfrac{3}{x-3}=\dfrac{3x-20}{\left(x-2\right)\left(x-3\right)}\left(x\ne2;x\ne3\right)\)

\(\Leftrightarrow\dfrac{2\left(x-3\right)}{\left(x-2\right)\left(x-3\right)}+\dfrac{3\left(x-2\right)}{\left(x-2\right)\left(x-3\right)}=\dfrac{3x-20}{\left(x-2\right)\left(x-3\right)}\)

\(\Leftrightarrow2\left(x-3\right)+3\left(x-2\right)=3x-20\)

\(\Leftrightarrow2x-6+3x-6=3x-20\)

\(\Leftrightarrow5x-12=3x-20\)

\(\Leftrightarrow5x-3x=-20+12\)

\(\Leftrightarrow2x=-8\)

\(\Leftrightarrow x=-4\left(tm\right)\)

Vậy: ...  

c) \(\dfrac{3}{x-2}+\dfrac{2}{x+1}=\dfrac{2x+5}{\left(x-2\right)\left(x+1\right)}\left(x\ne2;x\ne-1\right)\)

\(\Leftrightarrow\dfrac{3\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}+\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+1\right)}=\dfrac{2x+5}{\left(x-2\right)\left(x+1\right)}\)

\(\Leftrightarrow3\left(x+1\right)+2\left(x-2\right)=2x+5\)

\(\Leftrightarrow3x+3+2x-4=2x+5\)

\(\Leftrightarrow5x-1=2x+5\)

\(\Leftrightarrow5x-2x=5+1\)

\(\Leftrightarrow3x=6\)

\(\Leftrightarrow x=2\left(ktm\right)\)

Vậy pt vô nghiệm

\(\dfrac{1}{2x-3}=\dfrac{3}{2x^2-3x}+\dfrac{x}{5}\) (1)

 ĐK: \(\left\{{}\begin{matrix}2x-3\ne0\\2x^2-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{3}{2}\\x\ne0\end{matrix}\right.\)

(1) \(\Leftrightarrow\dfrac{1}{2x-3}=\dfrac{3}{x\left(2x-3\right)}+\dfrac{x}{5}\)

\(\Leftrightarrow\dfrac{5x}{5x\left(2x-3\right)}=\dfrac{15}{5x\left(2x-3\right)}+\dfrac{x^2\left(2x-3\right)}{5x\left(2x-3\right)}\)

\(\Leftrightarrow5x=15+x^2\left(2x-3\right)\)

\(\Leftrightarrow5x=15+2x^3-3x^2\)

\(\Leftrightarrow2x^3-3x^2-5x+15=0\) 

\(\Leftrightarrow x\approx-1,898\) (sử dụng máy tính) 

Vậy: ...