a: \(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\)

=>1+3x-6=3-x

=>3x-5=3-x

=>4x=8

hay x=2(loại)

b: \(\Leftrightarrow8-x-8\left(x-7\right)=-26\)

=>8-x-8x+56=-26

=>-9x+64=-26

=>-9x=-90

hay x=10(nhận)

c: \(\dfrac{1}{x-2}+\dfrac{1}{x-3}=\dfrac{2}{x-1}\)

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

\(\Leftrightarrow\left(x-1\right)\left(2x-5\right)=2\left(x^2-5x+6\right)\)

\(\Leftrightarrow2x^2-5x-2x+5=2x^2-10x+12\)

=>-7x+10x=12-5

=>3x=7

hay x=7/3(nhận)

Bài 1: 

a) \(-5\left(x^2-3x+1\right)+x\left(1+5x\right)=x-2\)

\(\Rightarrow-5x^2+15x-5+x+5x^2=x-2\)

\(\Rightarrow16x-5=x-2\)

\(\Rightarrow16x-x=5-2\)

\(\Rightarrow15x=3\)

\(\Rightarrow x=\dfrac{15}{3}=5\)

b) \(12x^2-4x\left(3x+5\right)=10x-17\)

\(\Rightarrow12x^2-12x^2-20x=10x-17\)

\(\Rightarrow-20x=10x-17\)

\(\Rightarrow-20x-10x=-17\)

\(\Rightarrow-30x=-17\)

\(\Rightarrow x=\dfrac{-30}{-17}=\dfrac{30}{17}\)

c) \(-4x\left(x-5\right)+7x\left(x-4\right)-3x^2=12\)

\(\Rightarrow-4x^2+20x+7x^2-28x-3x^2=12\)

\(\Rightarrow-8x=12\)

\(\Rightarrow x=\dfrac{12}{-8}=-\dfrac{4}{3}\)

Bài 2: 

a) \(\left(x+5\right)\left(x-7\right)-7x\left(x-3\right)\)

\(=x^2-7x+5x-35-7x^2+21x\)

\(=-6x^2+19x-35\)

b) \(x\left(x^2-x-2\right)-\left(x-5\right)\left(x+1\right)\)

\(=x^3-x^2-2x-x^2+x-5x-5\)

\(=x^3-2x^2-6x-5\)

c) \(\left(x-5\right)\left(x-7\right)-\left(x+4\right)\left(x-3\right)\)

\(=x^2-7x-5x+35-x^2-3x+4x-12\)

\(=11x+23\)

d) \(\left(x-1\right)\left(x-2\right)-\left(x+5\right)\left(x+2\right)\)

\(=x^2-2x-x+2-x^2+2x+5x+10\)

\(=4x+12\)

A = x(3x - 5) - (3x² - 7)

= 3x² - 5x - 3x² + 7

= -5x + 7

B = (x - 5)(x + 5) + x(3 - x)

= x² - 25 + 3x - x²

= 3x - 25

C = (x + 2)³ - (x - 2)³

= x³ + 3.x².2 + 3.x.2² + 2³ - (x³ - 3.x².2 + 3.x.2² - 2³)

= x³ + 6x² + 12x + 8 - x³ + 6x² - 12x + 8

= 12x² + 16

= 4(3x² + 4)

Câu 3:

x²-3x-4=0

\(\Leftrightarrow\)(x-4)(x+1)=0

\(\Leftrightarrow\)x=4 hoặc x=-1

Vậy...

`C(x)=`\(5-8x^4+2x^3+x+5x^4+x^2-4x^3\)

`C(x)= (-8x^4+5x^4)+(2x^3-4x^3)+x^2+x+5`

`C(x)= -3x^4-2x^3+x^2+x+5`

`D(x)=`\(\left(3x^5+x^4-4x\right)-\left(4x^3-7+2x^4+3x^5\right)\)

`D(x)= 3x^5+x^4-4x-4x^3+7-2x^4-3x^5`

`D(x)=(3x^5-3x^5)+(x^4-2x^4)-4x^3-4x+7`

`D(x)=-x^4-4x^3-4x+7`

`P(x)=C(x)+D(x)`

`P(x)=( -3x^4-2x^3+x^2+x+5)+(-x^4-4x^3-4x+7)`

`P(x)=-3x^4-2x^3+x^2+x+5-x^4-4x^3-4x+7`

`P(x)=(-3x^4-x^4)+(-2x^3-4x^3)+x^2+(x-4x)+(5+7)`

`P(x)=-4x^4-6x^3+x^2-3x+12`

`Q(x)=C(x)-D(x)`

`Q(x)=( -3x^4-2x^3+x^2+x+5)-(-x^4-4x^3-4x+7)`

`Q(x)=-3x^4-2x^3+x^2+x+5+x^4+4x^3+4x-7`

`Q(x)=(-3x^4+x^4)+(-2x^3+4x^3)+x^2+(x+4x)+(5-7)`

`Q(x)=-2x^4+2x^3+x^2+5x-2`

`F(x)=Q(x)-(-2x^4+2x^3+x^2-12)`

`F(x)=(-2x^4+2x^3+x^2+5x-2)-(-2x^4+2x^3+x^2-12)`

`F(x)=-2x^4+2x^3+x^2+5x-2+2x^4-2x^3-x^2+12`

`F(x)=(-2x^4+2x^4)+(2x^3-2x^3)+(x^2-x^2)+5x+(-2+12)`

`F(x)=5x+10`

Đặt `5x+10=0`

`\Leftrightarrow 5x=0-10`

`\Leftrightarrow 5x=-10`

`\Leftrightarrow x=-10 \div 5`

`\Leftrightarrow x=-2`

Vậy, nghiệm của đa thức là `x=-2.`

a: ĐKXĐ: \(x\notin\left\{3;-5\right\}\)

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

=>\(\dfrac{5\left(x+5\right)-3\left(x-3\right)}{15}=\dfrac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)

=>\(\dfrac{5x+25-3x+9}{15}=\dfrac{5x+25-3x+9}{\left(x-3\right)\left(x+5\right)}\)

=>(x-3)(x+5)=15

=>\(x^2+2x-15-15=0\)

=>\(x^2+2x-30=0\)

=>\(\left(x+1\right)^2=31\)

=>\(\left[{}\begin{matrix}x+1=\sqrt{31}\\x+1=-\sqrt{31}\end{matrix}\right.\Leftrightarrow x=-1\pm\sqrt{31}\left(nhận\right)\)

b: ĐKXĐ: \(x\in R\)

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

=>\(\left\{{}\begin{matrix}x^2+x+1=\left(3-x\right)^2\\x< =3\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =3\\x^2-6x+9=x^2+x+1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< =3\\-7x=-8\end{matrix}\right.\Leftrightarrow x=\dfrac{8}{7}\left(nhận\right)\)

c:

ĐKXĐ: \(x\in R\)

 \(x^2-x+\sqrt{x^2-x+24}=18\)

=>\(x^2-x+24+\sqrt{x^2-x+24}=42\)

=>\(\left(\sqrt{x^2-x+24}\right)^2+\left(\sqrt{x^2-x+24}\right)-42=0\)

=>\(\left(\sqrt{x^2-x+24}+7\right)\left(\sqrt{x^2-x+24}-6\right)=0\)

=>\(\sqrt{x^2-x+24}-6=0\)

=>\(x^2-x+24=36\)

=>\(x^2-x-12=0\)

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

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

a) 3 x 5 < 3 x 6

3 x 5 > 3x 4

b) 3 x 5 = 5 x 3

4 x 6 = 6 x 4

c) 20 : 4 > 20 : 5

20 : 4 < 20 : 2

a, 3 x 5  < 3 x 6

b, 3 x 5  > 3 x 4

c, 

a) 3 x 5 < 3 x 6; 3 x 5 > 3 x 4.
b) 3 x 5 = 5 x 3; 4 x 6 = 6 x 4.
c) 20 : 4 > 20 : 5; 20 : 4 < 20 : 2
 Mong em làm bài được điểm 10
 

1) \(5^{x+1}-5^x=20\Leftrightarrow5^x\left(5-1\right)=20\Leftrightarrow5^x=5\Leftrightarrow x=1\)

2) \(2^x+2^{x+4}=544\Leftrightarrow2^x\left(1+2^4\right)=544\Leftrightarrow2^x=32\Leftrightarrow x=5\)

3) \(4^{2x+1}+4^{2x}=80\Leftrightarrow4^{2x}\left(4+1\right)=80\Leftrightarrow16^x=16\Leftrightarrow x=1\)

4) \(3^{2x+2}+3^{2x+1}=108\Leftrightarrow3^{2x}\left(3^2+3\right)=108\Leftrightarrow9^x=9\Leftrightarrow x=1\)

5) \(7^{x+3}-7^{x+1}=16464\Leftrightarrow7^x\left(7^3-7\right)=16464\Leftrightarrow7^x=49\Leftrightarrow x=2\)

c:Ta có: \(5^{x+1}-5^x=20\)

\(\Leftrightarrow5^x\cdot5-5^x=20\)

\(\Leftrightarrow5^x\cdot4=20\)

\(\Leftrightarrow5^x=5\)

hay x=1

c: Ta có: \(2^x+2^{x+4}=544\)

\(\Leftrightarrow2^x+2^x\cdot16=544\)

\(\Leftrightarrow2^x\cdot17=544\)

\(\Leftrightarrow2^x=32\)

hay x=5

c: Ta có: \(4^{2x+1}+4^{2x}=80\)

\(\Leftrightarrow16^x\cdot4+16^x=80\)

\(\Leftrightarrow16^x\cdot5=80\)

\(\Leftrightarrow16^x=16\)

hay x=1

c: Ta có: \(3^{2x+2}+3^{2x+1}=108\)

\(\Leftrightarrow9^x\cdot9+9^x\cdot3=108\)

\(\Leftrightarrow9^x\cdot12=108\)

\(\Leftrightarrow9^x=9\)

hay x=1

c: Ta có: \(7^{x+3}-7^{x+1}=16464\)

\(\Leftrightarrow7^x\cdot343-7^x\cdot7=16464\)

\(\Leftrightarrow7^x\cdot336=16464\)

\(\Leftrightarrow7^x=49\)

hay x=2