a.\(2^x-2^4.2^7.32=0\)
\(2^x-2^{16}=0\)
\(=>x=16\)
b.\(3^x+3^{x+2}=270\)
\(3^x+3^x.3^2=270\)
\(3^x.10=270\)
\(3^x=27\)
\(=>x=3\)
 

\(a)\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{(x-3)^2(2x+5)}{(3x-1)(x-3)^2}(ĐK:x\ne3,x\ne\frac{1}{3})\)

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

Còn bài b bạn tự làm nhé

Điều kiện: \(x\ne\left\{-1;-2;-5\right\}\)

\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\frac{x^2\left(x+1\right)-4\left(x+1\right)}{x^2\left(x+1\right)+7x\left(x+1\right)+10\left(x+1\right)}\)

\(=\frac{\left(x+1\right)\left(x^2-4\right)}{\left(x+1\right)\left(x^2+7x+10\right)}\)

\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left[x\left(x+2\right)+5\left(x+2\right)\right]}\)

\(=\frac{\left(x+1\right)\left(x-2\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)\left(x+5\right)}=\frac{x-2}{x+5}\)

Điều kiện: \(x\ne\left\{3;\frac{1}{3}\right\}\)

\(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{2x^3-6x^2-x^2+3x-15x+45}{3x^3-9x^2-10x^2+30x+3x-9}\)

\(=\frac{2x^2\left(x-3\right)-x\left(x-3\right)-15\left(x-3\right)}{3x^2\left(x-3\right)-10x\left(x-3\right)+3\left(x-3\right)}\)

\(=\frac{\left(x-3\right)\left(2x^2-x-15\right)}{\left(x-3\right)\left(3x^2-10x+3\right)}\)

\(=\frac{2x^2-x-15}{3x^2-10x+3}=\frac{2x\left(x-3\right)+5\left(x-3\right)}{3x\left(x-3\right)-\left(x-3\right)}\)

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

a/ \(\dfrac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\dfrac{2x^3-12x^2+18x+5x^2-30x+45}{3x^3-18x^2+27x-x^2+6x-9}\)

\(=\dfrac{2x\left(x^2-6x+9\right)+5\left(x^2-6x+9\right)}{3x\left(x^2-6x+9\right)-\left(x^2-6x+9\right)}=\dfrac{\left(2x+5\right)\left(x^2-6x+9\right)}{\left(3x-1\right)\left(x^2-6x+9\right)}\)

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

b/ \(\dfrac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\dfrac{x^3+3x^2+2x-2x^2-6x-4}{x^3+3x^2+2x+5x^2+15x+10}\)

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

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

Lời giải:

ĐKXĐ:.........

a) \(\frac{2x^3-7x^2-12x+45}{3x^3-19x^2+33x-9}=\frac{2x^3-6x^2-(x^2-3x)-(15x-45)}{3x^3-9x^2-(10x^2-30x)+(3x-9)}\)

\(=\frac{2x^2(x-3)-x(x-3)-15(x-3)}{3x^2(x-3)-10x(x-3)+3(x-3)}=\frac{(x-3)(2x^2-x-15)}{(x-3)(3x^2-10x+3)}\)

\(=\frac{(x-3)[2x(x-3)+5(x-3)]}{(x-3)[3x(x-3)-(x-3)]}=\frac{(x-3)(x-3)(2x+5)}{(x-3)(x-3)(3x-1)}=\frac{2x+5}{3x-1}\)

b)

\(\frac{x^3+x^2-4x-4}{x^3+8x^2+17x+10}=\frac{x^2(x+1)-4(x+1)}{x^3+x^2+7x^2+7x+10x+10}\)

\(=\frac{(x+1)(x^2-4)}{x^2(x+1)+7x(x+1)+10(x+1)}=\frac{(x+1)(x-2)(x+2)}{(x+1)(x^2+7x+10)}\)

\(=\frac{(x-2)(x+2)}{x^2+7x+10}=\frac{(x-2)(x+2)}{x(x+2)+5(x+2)}=\frac{(x-2)(x+2)}{(x+2)(x+5)}=\frac{x-2}{x+5}\)

xin lỗi nhé mình mới có lớp 6 à nên ko bít

tha lỗi cho mình nhé!

\(\left(x^2+22x-120\right)\left(x^2+33x+270\right)-2x^2\)

\(=x^4+55x^3+876x^2+1980x-32400-2x^2\)

\(=x^4+55x^3+874x^2+1980x-32400\)

\(\left(x^2+22x+120\right)\left(x^2+33x+270\right)-2x^2\)

\(=\left(x+10\right)\left(x+12\right)\left(x+18\right)\left(x+15\right)-2x^2\)

\(=\left(x^2+28x+180\right)\left(x^2+27x+280\right)-2x^2\)

\(=\left(x^2+180\right)^2+55x\left(x^2+180\right)+754x^2\)

\(=\left(x^2+29x+180\right)\left(x^2+26x+180\right)\)

\(=\left(x+9\right)\left(x+20\right)\left(x^2+26x+180\right)\)

a: \(=\dfrac{6x^2+9x+8x+12}{2x+3}=\dfrac{3x\left(2x+3\right)+4\left(2x+3\right)}{2x+3}\)

=3x+4

b: \(=\dfrac{5x^2-2x+15x-6}{5x-2}\)

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

c: \(=\dfrac{-8x^2+20x+2x-5-10}{2x-5}=-4x+1+\dfrac{-10}{2x-5}\)

d: \(=\dfrac{14x^2-35x+2x-5}{2x-5}=\dfrac{7x\left(2x-5\right)+\left(2x-5\right)}{2x-5}\)

=7x+1

e: \(=\dfrac{2x^3+x^2+6x^2+3x+12x+6}{2x+1}\)

\(=\dfrac{x^2\left(2x+1\right)+3x\left(2x+1\right)+6\left(2x+1\right)}{2x+1}=x^2+3x+6\)

f: \(=\dfrac{x^3-2x^2+6x^2-12x+x-2}{x-2}=x^2+6x+1\)

g: \(=\dfrac{12x^3+6x^2-4x^2-2x+6x+3}{2x+1}=6x^2-2x+3\)

a/ 60x45-60x15-45x60+45x15

=(60x45-60x45)-(60x15-45x15)=0-15x(60-45)=-15x15=-225

b/68x12+68x32-32x68-12=68x12-12=12x67=804

c/-17x33+17x5+33x17-33x5

=5x(17-33)=5x(-16)=-80

a.

\(10-2\left(4-3x\right)=-4\)

\(\Leftrightarrow2\left(4-3x\right)=10+4\)

\(\Leftrightarrow2\left(4-3x\right)=14\)

\(\Leftrightarrow4-3x=7\)

\(\Leftrightarrow3x=-3\)

\(\Leftrightarrow x=-1\)

b.

\(-12+3\left(-x+7\right)=-18\)

\(\Leftrightarrow3\left(-x+7\right)=-18+12=-6\)

\(\Leftrightarrow-x+7=-6:3=-2\)

\(\Leftrightarrow x=9\)

c.

\(-45:5.\left(-3-2x\right)=3\)

\(\Leftrightarrow-9.\left(-3-2x\right)=3\)

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

\(\Leftrightarrow2x=-\dfrac{8}{3}\)

\(\Leftrightarrow x=-\dfrac{4}{3}\notin Z\left(loại\right)\)

Câu này em ghi sai đề?

d.

\(3x-28=x+36\)

\(\Leftrightarrow2x=28+36\)

\(\Leftrightarrow2x=64\)

\(\Leftrightarrow x=32\)

e.

\(\left(-12\right)^2.x=56+10.13x\)

\(\Leftrightarrow144x=56+130x\)

\(\Leftrightarrow144x-130x=56\)

\(\Leftrightarrow14x=56\)

\(\Leftrightarrow x=4\)

6.x - 37 = 1818 : 18                                           x : 23 + 45 = 270

6.x - 37 = 101                                                   x : 23         = 270 - 45

6.x        = 101 + 37                                           x : 23          = 225

6.x        = 138                                                        x           = 225 x 23

   x        = 138 : 6                                                   x           = 5175

   x        = 23

a) \(6x-37=1818:18\)

\(\Rightarrow6x-37=101\)

\(\Rightarrow6x=101+37\)

\(\Rightarrow6x=138\)

\(\Rightarrow x=138:6=23\)

b) \(x:23+45=270\)

\(x:23=270-45\)

\(x:23=225\)

\(x=225\cdot23\)

\(x=5175\)

a) 6.x - 37 = 1818 : 18

    6.x - 37 = 101

    6.x        = 101+ 37

    6.x        = 138

    x           = 138 : 6

    x           = 23