`1)(x+2)(x+3)(x-7)(x-8)=144`
`<=>[(x+2)(x-7)][(x+3)(x-8)]=144`
`<=>(x^2-5x-14)(x^2-5x-24)=144`
`<=>(x^2-5x-19)^2-25=144`
`<=>(x^2-5x-19)^2-169=0`
`<=>(x^2-5x-6)(x^2-5x-32)=0`
`+)x^2-5x-6=0`
`<=>` $\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.$
`+)x^2-5x-32=0`
`<=>` $\left[ \begin{array}{l}x=\dfrac{5+3\sqrt{17}}{2}\\x=\dfrac{5-3\sqrt{17}}{2}\end{array} \right.$
Vậy `S={-1,6,\frac{5+3\sqrt{17}}{2},\frac{5-3\sqrt{17}}{2}}`

1: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)=144\)

\(\Leftrightarrow\left(x^2-7x+2x-14\right)\left(x^2-8x+3x-24\right)=144\)

\(\Leftrightarrow\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2-38\left(x^2-5x\right)+336-144=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2-38\left(x^2-5x\right)+192=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2-6\left(x^2-5x\right)-32\left(x^2-5x\right)+192=0\)

\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x-6\right)-32\left(x^2-5x-6\right)=0\)

\(\Leftrightarrow\left(x^2-5x-6\right)\left(x^2-5x-32\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+1\right)\left(x^2-5x-32\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+1=0\\x^2-5x-32=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\\x=\dfrac{5-3\sqrt{17}}{2}\\x=\dfrac{5+3\sqrt{17}}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{6;-1;\dfrac{5-3\sqrt{17}}{2};\dfrac{5+3\sqrt{17}}{2}\right\}\)

`2)(6x+5)^2(3x+2)(x+1)=35`
`<=>12(6x+5)^2(3x+2)(x+1)=420`
`<=>(6x+5)^2+(6x+4)(6x+6)=420`
Đặt `6x+5=a` 
`pt<=>a^2(a+1)(a-1)=420`
`<=>a^2(a^2-1)-420=0`
`<=>a^4-a^2-420=0`
`<=>` $\left[ \begin{array}{l}a^2=-20(False)\\a^2=21(True)\end{array} \right.$
`<=>` $\left[ \begin{array}{l}a=\sqrt{20}\\a=-\sqrt{20}\end{array} \right.$
`<=>` $\left[ \begin{array}{l}6x+5=\sqrt{20}\\6x+5=-\sqrt{20}\end{array} \right.$
`<=>` $\left[ \begin{array}{l}x=\dfrac{\sqrt{20}-5}{6}\\x=\dfrac{-\sqrt{20}-5}{6}\end{array} \right.$
Vậy `S={\frac{\sqrt{20}-5}{6},\frac{-\sqrt{20}-5}{6}}`

\(\left(\dfrac{2}{5}-\dfrac{1}{4}\right):\dfrac{3}{4}x\dfrac{5}{2}=\dfrac{3}{20}:\dfrac{3}{4}x\dfrac{5}{2}=\dfrac{3}{20}x\dfrac{4}{3}x\dfrac{5}{2}=\dfrac{3x4x5}{20x3x2}=\dfrac{1}{2}\)

`( 2 / 5 - 1 / 4 ) : 3 / 4 xx 5 / 2`

`=( 8 / 20 - 5 / 20 ) xx 4 / 3 xx 5 / 2`

`= 3 / 20 xx 4 / 3 xx 5 / 2`

`= [ 3 xx 4 xx 5 ] / [ 4 xx 5 xx 3 xx 2 ]`

`= 1 / 2`

(2/5 - 1/4) : 3/4 x 5/2
= 3/20 : 3/4 x 5/2
= 3/20 x 4/3 x 5/2 ( chia là nhân nghịch đảo)
= 1/5 x 5/2
=1/2 hay 0,5

Mọi người đánh giúp mình nhé! Hạn là tối nay!!

Lời giải:

8A=(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$

$=[(3^2-1)(3^2+1)](3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$

$=(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)-4.3^{32}$

$=(3^8-1)(3^8+1)(3^{16}+1)-4.3^{32}$

$=(3^{16}-1)(3^{16}+1)-4.3^{32}$

$=3^{32}-1-4.3^{32}$

$=-3.3^{32}-1=-3^{33}-1$
$\Rightarrow A=\frac{-3^{33}-1}{8}$

\(a,\dfrac{1}{2}+\left(-\dfrac{3}{4}\right)=\dfrac{2}{4}+\left(-\dfrac{3}{4}\right)=-\dfrac{1}{4}\)

\(b,\dfrac{2}{5}+\left(-\dfrac{1}{16}\right)+\left(-\dfrac{9}{5}\right)+\left(-\dfrac{5}{16}\right)=\left(\dfrac{2}{5}-\dfrac{9}{5}\right)-\left(\dfrac{1}{16}+\dfrac{5}{16}\right)=\dfrac{-7}{5}-1=\dfrac{-12}{5}\)

\(c,\dfrac{-2}{13}+\dfrac{8}{15}+\dfrac{10}{20}+\dfrac{15}{13}+\left(-\dfrac{23}{15}\right)=\left(\dfrac{-2}{13}+\dfrac{15}{13}\right)+\left(\dfrac{8}{15}+\dfrac{-23}{15}\right)+\dfrac{1}{2}=1+\left(-1\right)+\dfrac{1}{2}=\dfrac{1}{2}\)

(Cho mik làm lại nha)

\(a,\dfrac{1}{2}+\left(-\dfrac{3}{4}\right)=\dfrac{2}{4}+\left(-\dfrac{3}{4}\right)=-\dfrac{1}{4}\)

\(a,\dfrac{2}{5}+\left(-\dfrac{1}{16}\right)+\left(-\dfrac{9}{5}\right)+\left(-\dfrac{5}{16}\right)=\left(\dfrac{2}{5}+\dfrac{-9}{5}\right)-\left(\dfrac{1}{16}+\dfrac{5}{16}\right)=\dfrac{-7}{5}-\dfrac{6}{16}=\dfrac{-71}{40}\)

\(c,\dfrac{-2}{13}+\dfrac{8}{15}+\dfrac{10}{20}+\dfrac{15}{13}+\left(-\dfrac{23}{15}\right)=\left(\dfrac{-2}{13}+\dfrac{15}{13}\right)+\left(\dfrac{8}{15}+\dfrac{-23}{15}\right)+\dfrac{1}{2}=1+\left(-1\right)+\dfrac{1}{2}=\dfrac{1}{2}\)

a, 16/2n=2

<=>2n=8

<=>n=4

b, (-3)^n =-27*81=-2187

n=7( vì (-3)^7 =-2187

c, 8^n : 2^n =4

<=> (8:2)^n=4

4^n=4

n=1

Câu 1: \(\frac{4^9.36+64^4}{16^4.100}=\frac{4^9.36+4^{12}}{4^8.4.5^2}=\frac{4^9\left(36+4^3\right)}{4^9.25}=\frac{4^9.100}{4^9.25}=\frac{100}{25}=4\)

Câu 2: \(\frac{125^5-25^5}{100.50^5-10^2}=\frac{25^5.5^5-25^5}{25.4.25^5.2^5-25.4}=\frac{25^5\left(5^5-1\right)}{100\left(25^5.2^5-1\right)}\)

Câu 2 sai đề hay sao á bạn mk làm ko ra!