2:

a: =>x-4>=0

=>x>=4

b: =>x+1>0

=>x>-1

c.

\(\Leftrightarrow cos\left(x+12^0\right)+cos\left(90^0-78^0+x\right)=1\)

\(\Leftrightarrow2cos\left(x+12^0\right)=1\)

\(\Leftrightarrow cos\left(x+12^0\right)=\dfrac{1}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+12^0=60^0+k360^0\\x+12^0=-60^0+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=48^0+k360^0\\x=-72^0+k360^0\end{matrix}\right.\)

2.

Do \(-1\le sin\left(3x-27^0\right)\le1\) nên pt có nghiệm khi:

\(\left\{{}\begin{matrix}2m^2+m\ge-1\\2m^2+m\le1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m+1\ge0\left(luôn-đúng\right)\\2m^2+m-1\le0\end{matrix}\right.\)

\(\Rightarrow-1\le m\le\dfrac{1}{2}\)

a.

\(\Rightarrow\left[{}\begin{matrix}x+15^0=arccos\left(\dfrac{2}{5}\right)+k360^0\\x+15^0=-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-15^0+arccos\left(\dfrac{2}{5}\right)+k360^0\\x=-15^0-arccos\left(\dfrac{2}{5}\right)+k360^0\end{matrix}\right.\)

b.

\(2x-10^0=arccot\left(4\right)+k180^0\)

\(\Rightarrow x=5^0+\dfrac{1}{2}arccot\left(4\right)+k90^0\)

2.

Phương trình \(sin\left(3x-27^o\right)=2m^2+m\) có nghiệm khi:

\(2m^2+m\in\left[-1;1\right]\)

\(\Leftrightarrow\left\{{}\begin{matrix}2m^2+m\le1\\2m^2+m\ge-1\end{matrix}\right.\)

\(\Leftrightarrow\left(m+1\right)\left(2m-1\right)\le0\)

\(\Leftrightarrow-1\le m\le\dfrac{1}{2}\)

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Bài 1.

a) ( x - 3 )( x + 7 ) = 0

<=> x - 3 = 0 hoặc x + 7 = 0

<=> x = 3 hoặc x = -7

Vậy S = { 3 ; -7 }

b) ( x - 2 )² + ( x - 2 )( x - 3 ) = 0

<=> ( x - 2 )( x - 2 + x - 3 ) = 0

<=> ( x - 2 )( 2x - 5 ) = 0

<=> x - 2 = 0 hoặc 2x - 5 = 0

<=> x = 2 hoặc x = 5/2

Vậy S = { 2 ; 5/2 }

c) x² - 5x + 6 = 0

<=> x² - 2x - 3x + 6 = 0

<=> x( x - 2 ) - 3( x - 2 ) = 0

<=> ( x - 2 )( x - 3 ) = 0

<=> x - 2 = 0 hoặc x - 3 = 0

<=> x = 2 hoặc x = 3

Bài 2.

a) \(\frac{x}{x+1}-1=\frac{3}{2}x\)

ĐKXĐ : x khác -1

<=> \(\frac{x}{x+1}-\frac{x+1}{x+1}=\frac{3}{2}x\)

<=> \(\frac{-1}{x+1}=\frac{3x}{2}\)

=> 3x( x + 1 ) = -2

<=> 3x² + 3x + 2 = 0

Vi 3x² + 3x + 2 = 3( x² + x + 1/4 ) + 5/4 = 3( x + 1/2 )² + 5/4 ≥ 5/4 > 0 ∀ x

=> phương trình vô nghiệm

b) \(\frac{4x}{x-2}-\frac{7}{x}=4\)

ĐKXĐ : x khác 0 ; x khác 2

<=> \(\frac{4x^2}{x\left(x-2\right)}-\frac{7x-14}{x\left(x-2\right)}=\frac{4x^2-8x}{x\left(x-2\right)}\)

=> 4x² - 7x + 14 = 4x² - 8x

<=> 4x² - 7x - 4x² + 8x = -14

<=> x = -14 ( tm )

Vậy phương trình có nghiệm x = -14

a: \(\Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\)

=>(4x+14+3x+9)(4x+14-3x-9)=0

=>(7x+23)(x+5)=0

=>x=-23/7 hoặc x=-5

\(a,\\ \Leftrightarrow7x^2+58x+115=0\\ \Leftrightarrow\left(x+5\right)\left(7x+23\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x+5=0\\7x+23=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-5\\x=-\dfrac{23}{7}\end{matrix}\right.\)

\(b,\\ \Leftrightarrow\left[\left(x+1\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]=0\\ \Leftrightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=0\\ \LeftrightarrowĐặt.x^2+6x+5=a\\ \Leftrightarrow a=a\left(a+3\right)=10\\ \Leftrightarrow a^2+3a-10=0\\ \Leftrightarrow\left(a+5\right)\left(a-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}a=-5\\a=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x^2+6x+5=-5\\x^2+6x+5=2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2+6x+10=0\\x^2+6x+3=0\end{matrix}\right.\\ \left(Vô.n_o\Delta=36-40=-4< 0\right)\) 

\(\Leftrightarrow\left[{}\begin{matrix}x=-3+\sqrt{6}\\x=-3-\sqrt{6}\end{matrix}\right.\)

a: 7x+35=0

=>7x=-35

=>x=-5

b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)

=>8-x-8(x-7)=1

=>8-x-8x+56=1

=>-9x+64=1

=>-9x=-63

hay x=7(loại)

a, \(7x=-35\Leftrightarrow x=-5\)

b, đk : x khác 7 

\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)

vậy pt vô nghiệm 

2, thiếu đề 

1.

\(a,7x+35=0\\ \Rightarrow7x=-35\\ \Rightarrow x=-5\\ b,ĐKXĐ:x\ne7\\ \dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\\ \Leftrightarrow\dfrac{8-x}{x-7}-\dfrac{8\left(x-7\right)}{x-7}-\dfrac{1}{x-7}=0\\ \Leftrightarrow\dfrac{8-x-8x+56-1}{x-7}=0\\ \Rightarrow-9x+63=0\\ \Leftrightarrow-9x=-63\\ \Leftrightarrow x=7\left(ktm\right)\)

2.đề thiếu

ĐKXĐ: \(x\neq 0\).

Đặt \(\dfrac{x}{3}-\dfrac{4}{x}=t\).

PT đã cho tương đương:

\(3t^2+8-10t=0\)

\(\Leftrightarrow\left(t-2\right)\left(3t-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=\dfrac{4}{3}\end{matrix}\right.\).

Với t = 2 ta có \(\dfrac{x}{3}-\dfrac{4}{x}=2\Leftrightarrow\dfrac{x^2-12}{3x}=2\Leftrightarrow x^2-6x-12=0\Leftrightarrow x=\pm\sqrt{21}+3\).

Với t = \(\frac{4}{3}\) ta có \(\dfrac{x}{3}-\dfrac{4}{x}=\dfrac{4}{3}\Leftrightarrow\dfrac{x^2-12}{3x}=\dfrac{4}{3}\Leftrightarrow x^2-12=4x\Leftrightarrow x^2-4x-12=0\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-2\end{matrix}\right.\).

Vậy...

a/ 4x + 20 = 0

⇔4x = -20

⇔x = -5

Vậy phương trình có tập nghiệm S = {-5}

b/ 2x – 3 = 3(x – 1) + x + 2

⇔ 2x-3 = 3x -3+x+2

⇔2x – 3x = -3+2+3

⇔-2x = 2

⇔x = -1

Vậy phương trình có tập nghiệm S = {-1}
 

câu tiếp theo

a/ (3x – 2)(4x + 5) = 0

3x – 2 = 0 hoặc 4x + 5 = 0

Vậy phương trình có tập nghiệm S= {-5/4,3/2}

b/ 2x(x – 3) – 5(x – 3) = 0

=> (x – 3)(2x -5) = 0

=> x – 3 = 0 hoặc 2x – 5 = 0

* x – 3 = 0 => x = 3

* 2x – 5 = 0 => x = 5/2

Vậy phương trình có tập nghiệm S = {0, 5/2}

b1

a. 4x+ 20=0 <=> 4x= -20 <=> x= -20/4 <=> x= -5

b. 2x- 3= 3(x- 1)+ x+ 2 <=> 2x- 3= 3x- 3+ x+ 2

<=> 2x- 3= 4x- 1 <=> 2x- 4x= -1+ 3 <=> -2x= 2

<=> x= 2/-2 <=> x= -1

b2

a. (3x- 2)(4x+ 5)= 0

<=>\(\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=2\\4x=-5\end{cases}}}\)

<=>\(\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)

b. 2x(x- 3)- 5(x- 3)= 0

<=> (x- 3)(2x- 5)= 0

<=> \(\orbr{\begin{cases}x-3=0\\2x-5=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\2x=5\end{cases}}\)

<=> \(\orbr{\begin{cases}x=3\\x=\frac{5}{2}\end{cases}}\)

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

a) \(3x-2\sqrt{x-1}=4\) (ĐK: x ≥ 1)

\(\Rightarrow3x-2\sqrt{x-1}-4=0\)

\(\Rightarrow3x-6-2\sqrt{x-1}+2=0\)

\(\Rightarrow3\left(x-2\right)-2\left(\sqrt{x-1}-1\right)=0\)

\(\Rightarrow3\left(x-2\right)-2.\dfrac{x-2}{\sqrt{x-1}+1}=0\)

\(\Rightarrow\left(x-2\right)\left[3-\dfrac{2}{\sqrt{x-1}+1}\right]=0\)

*TH1: x = 2 (t/m)

*TH2: \(3-\dfrac{2}{\sqrt{x-1}+1}=0\)

\(\Rightarrow3=\dfrac{2}{\sqrt{x-1}+1}\)

\(\Rightarrow3\sqrt{x-1}+3=2\)

\(\Rightarrow3\sqrt{x-1}=-1\) (vô lí)

Vậy S = {2}

b) \(\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\) (ĐK: \(-\dfrac{1}{4}\le x\le3\) )

\(\Rightarrow\sqrt{4x+1}-3-\sqrt{x+2}+2-\sqrt{3-x}+1=0\)

\(\Rightarrow\dfrac{4x-8}{\sqrt{4x+1}+3}-\dfrac{x-2}{\sqrt{x+2}+2}+\dfrac{x-2}{\sqrt{3-x}+1}=0\)

\(\Rightarrow\left(x-2\right)\left(\dfrac{4}{\sqrt{4x+1}+3}-\dfrac{1}{\sqrt{x+2}+2}+\dfrac{1}{\sqrt{3-x}+1}\right)=0\)

=> x = 2

\(a,3x-2\sqrt{x-1}=4\left(x\ge1\right)\\ \Leftrightarrow-2\sqrt{x-1}=4-3x\\ \Leftrightarrow4\left(x-1\right)=16-24x+9x^2\\ \Leftrightarrow9x^2-28x+20=0\\ \Leftrightarrow\left(x-2\right)\left(9x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=\dfrac{10}{9}\left(tm\right)\end{matrix}\right.\)

\(b,\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\left(-\dfrac{1}{4}\le x\le3\right)\\ \Leftrightarrow4x+1+x+2-2\sqrt{\left(4x+1\right)\left(x+2\right)}=3-x\\ \Leftrightarrow-2\sqrt{\left(4x+1\right)\left(x+2\right)}=2-6x\\ \Leftrightarrow\sqrt{4x^2+9x+2}=3x-1\\ \Leftrightarrow4x^2+9x+2=9x^2-6x+1\\ \Leftrightarrow5x^2-15x-1=0\\ \Leftrightarrow\Delta=225+20=245\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15-\sqrt{245}}{10}=\dfrac{15-7\sqrt{5}}{10}\left(ktm\right)\\x=\dfrac{15+\sqrt{245}}{10}=\dfrac{15+7\sqrt{5}}{10}\left(tm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{15+7\sqrt{5}}{10}\)