\(2\left(x+1\right)=5x+7\\ \Leftrightarrow2x+2=5x+7\\\Leftrightarrow 2x-5x=-2+7\\\Leftrightarrow -3x=5\\ \Leftrightarrow x=-\frac{5}{3}\)

Vậy phương trình trên có nghiệm là \(-\frac{5}{3}\)

\(3x-1=x+3\\ \Leftrightarrow3x-x=1+3\\ \Leftrightarrow2x=4\\\Leftrightarrow x=2\)

Vậy phương trình trên có nghiệm là \(2\)

\(15-7x=9-3x\\\Leftrightarrow -7x+3x=-15+9\\\Leftrightarrow -4x=-6\\ \Leftrightarrow x=\frac{3}{2}\)

Vậy phương trình trên có nghiệm là \(\frac{3}{2}\)

\(2x+1=15x-5\\ \Leftrightarrow2x-15x=-1-5\\ \Leftrightarrow-13x=-6\\ \Leftrightarrow x=\frac{6}{13}\)

Vậy phương trình trên có nghiệm là \(\frac{6}{13}\)

\(3x-2=2x+5\\ \Leftrightarrow3x-2x=2+5\\ \Leftrightarrow x=7\)

Vậy phương trình trên có nghiệm là \(7\)

ta có : x^5+2x^4+3x^3+3x^2+2x+1=0

\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0

\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0

\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0

\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0

\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0

\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0

VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)

\(\Rightarrow\)x+1=0

\(\Rightarrow\)x=-1

CÒN CÂU B TỰ LÀM (02042006)

b: x^4+3x^3-2x^2+x-3=0

=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0

=>(x-1)(x^3+4x^2+2x+3)=0

=>x-1=0

=>x=1

1) Ta có: 3x-12=5x(x-4)

\(\Leftrightarrow3x-12-5x\left(x-4\right)=0\)

\(\Leftrightarrow3x-12-5x^2+20x=0\)

\(\Leftrightarrow-5x^2+23x-12=0\)

\(\Leftrightarrow-5x^2+20x+3x-12=0\)

\(\Leftrightarrow\left(-5x^2+20x\right)+\left(3x-12\right)=0\)

\(\Leftrightarrow5x\left(-x+4\right)+3\left(x-4\right)=0\)

\(\Leftrightarrow5x\left(4-x\right)-3\left(4-x\right)=0\)

\(\Leftrightarrow\left(4-x\right)\left(5x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\5x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\5x=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{3}{5}\end{matrix}\right.\)

Vậy: \(x\in\left\{4;\frac{3}{5}\right\}\)

2) Ta có: 3x-15=2x(x-5)

\(\Leftrightarrow3x-15-2x\left(x-5\right)=0\)

\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\end{matrix}\right.\)

Vậy: \(x\in\left\{5;\frac{3}{2}\right\}\)

3) Ta có: 3x(2x-3)+2(2x-3)=0

\(\Leftrightarrow\left(2x-3\right)\left(3x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=3\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{-2}{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{3}{2};-\frac{2}{3}\right\}\)

4) Ta có: (4x-6)(3-3x)=0

\(\Leftrightarrow\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=6\\3x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{6}{4}=\frac{3}{2}\\x=1\end{matrix}\right.\)

Vậy: \(x\in\left\{\frac{3}{2};1\right\}\)

4) (4x - 6 ) ( 3 - 3x ) = 0

<=> \(\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}4x=6\\3x=3\end{matrix}\right.\)

<=> \(\left[{}\begin{matrix}x=\frac{3}{2}\\x=1\end{matrix}\right.\)

Bài 1 :

a, Ta có : \(3x-12=5x\left(x-4\right)\)

=> \(3x-12=5x^2-20x\)

=> \(3x-12-5x^2+20x=0\)

=> \(5x^2-23x+12=0\)

=> \(5x^2-20x-3x+12=0\)

=> \(5x\left(x-4\right)-3\left(x-4\right)=0\)

=> \(\left(5x-3\right)\left(x-4\right)=0\)

=> \(\left[{}\begin{matrix}5x-3=0\\x-4=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{3}{5}\\x=4\end{matrix}\right.\)

Vậy phương trình có nghiệm là x = \(\frac{3}{5}\) và x = 4 .

b, Ta có : \(3x-15=2x\left(x-5\right)\)

=> \(3x-15-2x\left(x-5\right)=0\)

=> \(3\left(x-5\right)-2x\left(x-5\right)=0\)

=> \(\left(3-2x\right)\left(x-5\right)=0\)

=> \(\left[{}\begin{matrix}3-2x=0\\x-5=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{3}{2}\\x=5\end{matrix}\right.\)

Vậy phương trình có nghiệm là x = \(\frac{3}{2}\) và x = 5 .

c, Ta có : \(3x\left(2x-3\right)+2\left(2x-3\right)=0\)

=> \(\left(3x+2\right)\left(2x-3\right)=0\)

=> \(\left[{}\begin{matrix}3x+2=0\\2x-3=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}3x=-2\\2x=3\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=-\frac{2}{3}\\x=\frac{3}{2}\end{matrix}\right.\)

Vậy phương trình có nghiệm là x = \(-\frac{2}{3}\) và x = \(\frac{3}{2}\) .

d, Ta có : \(\left(4x-6\right)\left(3-3x\right)=0\)

=> \(\left[{}\begin{matrix}4x-6=0\\3-3x=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}4x=6\\-3x=-3\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{6}{4}\\x=1\end{matrix}\right.\)

Vậy phương trình có nghiệm là x = 1 và x = \(\frac{6}{4}\) .

a: =>2x-3x^2-x<15-3x^2-6x

=>x<-6x+15

=>7x<15

=>x<15/7

b: =>4x^2-24x+36-4x^2+4x-1>=12x

=>-20x+35>=12x

=>-32x>=-35

=>x<=35/32

\(a,2x-x\left(3x+1\right)< 15-3x\left(x+2\right)\\ \Leftrightarrow2x-3x^2-x< 15-3x^2-6x\\ \Leftrightarrow3x^2-3x^2+2x+6x-x< 15\\ \Leftrightarrow7x< 15\\ \Leftrightarrow x< \dfrac{15}{7}\)

Vậy S={-∞; 15/7}

\(b,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12x\\ \Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-4x+1\right)-12x\ge0\\ \Leftrightarrow4x^2-4x^2-24x+4x-12x\ge-36+1\\ \Leftrightarrow-32x\ge-35\\ \Leftrightarrow x\le\dfrac{35}{32}\)

Vậy S={-∞; 35/32]

a, đk : x >= 1

\(\left[{}\begin{matrix}3x+5=2x-2\\3x+5=2-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\\x=-\dfrac{3}{5}\end{matrix}\right.\left(ktm\right)\)

vậy pt vô nghiệm 

b, đk >= 0 

\(\left[{}\begin{matrix}x^2+1=2x\\x^2+1=-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

c, \(\left[{}\begin{matrix}2x^2+2x=0\\2x^2+4x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x\left(x+1\right)=0\\x^2+2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0;x=-1\\x=-1\end{matrix}\right.\)

a) Để biểu thức vô nghĩa thì \(\dfrac{3x-2}{5}-\dfrac{x-4}{3}=0\)

\(\Leftrightarrow\dfrac{3x-2}{5}=\dfrac{x-4}{3}\)

\(\Leftrightarrow3\left(3x-2\right)=5\left(x-4\right)\)

\(\Leftrightarrow9x-6=5x-20\)

\(\Leftrightarrow9x-5x=-20+6\)

\(\Leftrightarrow4x=-14\)

\(\Leftrightarrow x=-\dfrac{7}{2}\)

a.ĐKXĐ:\(\frac{3}{2}\le x\le\frac{5}{2}\)

AD BĐT Cauchy ta được:

\(\sqrt{\left(2x-3\right)1}\le\frac{2x-3+1}{2}=\frac{2x-2}{2}=x-1\)

\(\sqrt{\left(5-2x\right)\cdot1}\le\frac{5-2x+1}{2}=\frac{6-2x}{2}=3-x\)

Do đó \(\sqrt{2x-3}+\sqrt{5-2x}\le x-1+3-x=2\)(1)

Lại có \(3x^2-12x+14=3\left(x-2\right)^2+2\ge2\)(2)

Từ (1) và (2) suy ra \(\sqrt{2x-3}+\sqrt{5-2x}\le3x^2-12x+14\)

Dấu = khi x=2 (tm ĐKXĐ)

PHẦN b giải tương tự

c.

ĐKXĐ: \(\left[{}\begin{matrix}x\le-5\\x\ge6\end{matrix}\right.\)

\(\sqrt{\left(x-3\right)\left(x-5\right)}+\sqrt{\left(x-3\right)\left(x+5\right)}=\sqrt{\left(x-3\right)\left(x-6\right)}\)

- Với \(x\ge6\) , do \(x-3>0\) pt trở thành:

\(\sqrt{x-5}+\sqrt{x+5}=\sqrt{x-6}\)

Do \(\left\{{}\begin{matrix}\sqrt{x-5}>\sqrt{x-6}\\\sqrt{x+5}>0\end{matrix}\right.\) \(\Rightarrow\sqrt{x-5}+\sqrt{x+5}>\sqrt{x-6}\) pt vô nghiệm

- Với \(x\le-5\) pt tương đương:

\(\sqrt{\left(3-x\right)\left(5-x\right)}+\sqrt{\left(3-x\right)\left(-x-5\right)}=\sqrt{\left(3-x\right)\left(6-x\right)}\)

Do \(3-x>0\) pt trở thành:

\(\sqrt{5-x}+\sqrt{-x-5}=\sqrt{6-x}\)

\(\Leftrightarrow-2x+2\sqrt{x^2-25}=6-x\)

\(\Leftrightarrow2\sqrt{x^2-25}=x+6\) (\(x\ge-6\))

\(\Leftrightarrow4\left(x^2-25\right)=x^2+12x+36\)

\(\Leftrightarrow3x^2-12x-136=0\Rightarrow x=\dfrac{6-2\sqrt{111}}{3}\)

a.

Kiểm tra lại đề, pt này không giải được

b.

ĐKXĐ: \(x\ge0\)

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

\(\Leftrightarrow\sqrt{x}\left(\sqrt{x+1}-1\right)-\left(\sqrt{x+1}-1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x+1}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x+1}=1\end{matrix}\right.\)

\(\Leftrightarrow...\)