x=2,3,4,5

\(25\leq5^x\leq3125\)

\(\implies 5^2\leq5^x\leq5^5\)

\(\implies 2\leq x\leq5\)

\(\implies x\in\{2;3;4;5\}\)

~ Học tốt a~

\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)

\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)

\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)

\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)

1a) (2x - 6)(x + 2) = 0

=> \(\orbr{\begin{cases}2x-6=0\\x+2=0\end{cases}}\)

=> \(\orbr{\begin{cases}2x=6\\x=-2\end{cases}}\)

=> \(\orbr{\begin{cases}x=3\\x=-2\end{cases}}\)

b) (x² + 7)(x² - 25) = 0

=> \(\orbr{\begin{cases}x^2+7=0\\x^2-25=0\end{cases}}\)

=> \(\orbr{\begin{cases}x^2=-7\\x^2=25\end{cases}}\)

=>  x ko có giá trị vì x² \(\ge\)0 mà x²= -7

hoặc x = \(\pm\)5

suy ra 2x-6 =0 hoặc x+2=0

sau đó bạn giải từng trường hợp

Bài 1:

a: ĐKXĐ: x<>0; x<>3; x<>1

\(A=\left(\frac{x-3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}\right):\frac{2x-2}{x}\)

\(=\frac{\left(x-3\right)^2-x^2+9}{x\left(x-3\right)}:\frac{2\left(x-1\right)}{x}\)

\(=\frac{x^2-6x+9-x^2+9}{x\cdot\left(x-3\right)}:\frac{2\left(x-1\right)}{x}=\frac{-6\left(x-3\right)}{x\left(x-3\right)}:\frac{2\left(x-1\right)}{x}=\frac{-6}{x}\cdot\frac{x}{2\left(x-1\right)}=\frac{-3}{x-1}\)

b: Để A là số nguyên thì -3⋮x-1

=>x-1∈{1;-1;3;-3}

=>x∈{2;0;4;-2}

Kết hợp ĐKXĐ, ta được: x∈{2;4;-2}

Bài 2:

a: \(x^3-2x^2=x^2\cdot x-x^2\cdot2=x^2\left(x-2\right)\)

b: \(y^2-2y-x^2+1\)

\(=\left(y^2-2y+1\right)-x^2\)

\(=\left(y-1\right)^2-x^2=\left(y-1-x\right)\left(y-1+x\right)\)

c: \(\left(x+1\right)^2-25\)

=(x+1+5)(x+1-5)

=(x-4)(x+6)

bn ... ơi...mik ...bỏ...cuộc ...hu...hu

. Huhu T^T mong sẽ có ai đó giúp mình "((

a)

\(4^x+3=32768\)

\(4^x=32768-3\)

\(4^x=32765\)

sai rồi ý mình ko phải thế để mình viết lại câu hỏi rồi bạn trả lời nhé.mà bạn học lớp mấy thế ? phải trả lời học lớp mấy nhé!

a: \(280-\left(x-140\right):35=270\)

=>(x-140):35=280-270=10

=>x-140=350

=>x=350+140

=>x=490

b: \(\left(190-2x\right):35-32=16\)

=>\(\left(190-2x\right):35=32+16=48\)

=>\(190-2x=35\cdot48=1680\)

=>2x=190-1680=-1490

=>x=-745

c: \(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=2^3\cdot5\)

=>\(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=8\cdot5=40\)

=>41-(2x-5)=720:40=18

=>2x-5=41-18=23

=>2x=28

=>x=14

d: \(\left(x:23+45\right)\cdot37-22=2^4\cdot105\)

=>\(\left(\frac{x}{23}+45\right)\cdot37=16\cdot105+22=1702\)

=>\(\frac{x}{23}+45=46\)

=>\(\frac{x}{23}=1\)

=>x=23

e: \(\left(3x-4\right)\left(x-1\right)^3=0\)

=>\(\left[\begin{array}{l}3x-4=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac43\\ x=1\end{array}\right.\)

f: \(2^{2x-1}:4=8^3\)

=>\(2^{2x-1-2}=2^9\)

=>2x-3=9

=>2x=12

=>x=6

g: \(x^{17}=x\)

=>\(x^{17}-x=0\)

=>\(x\left(x^{16}-1\right)=0\)

=>\(\left[\begin{array}{l}x=0\\ x^{16}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^{16}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-1\end{array}\right.\)

h: \(\left(x-5\right)^4=\left(x-5\right)^6\)

=>\(\left(x-5\right)^6-\left(x-5\right)^4=0\)

=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)

=>\(\left(x-5\right)^4\cdot\left(x-4\right)\left(x-6\right)=0\)

=>x∈{4;5;6}

i: \(\left(x+2\right)^5=2^{10}\)

=>\(\left(x+2\right)^5=\left(2^2\right)^5=4^5\)

=>x+2=4

=>x=2

k: 1+2+3+...+x=78

=>\(\frac{x\left(x+1\right)}{2}=78\)

=>x(x+1)=156

=>\(x^2+x-156=0\)

=>(x+13)(x-12)=0

=>x=-13(loại) hoặc x=12(nhận)

l: \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)

=>\(3x-16=2\cdot\frac{7^4}{7^3}=2\cdot7=14\)

=>3x=16+14=30

=>\(x=\frac{30}{3}=10\)

n: \(5^{x}:5^2=125\)

=>\(5^{x-2}=5^3\)

=>x-2=3

=>x=5

m: \(\left(x+1\right)^2=\left(x+1\right)^0\)

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

=>\(\left[\begin{array}{l}x+1=1\\ x+1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-2\end{array}\right.\)

o: Số số hạng của dãy số 2;4;..;52 là:

(52-2):2+1=50:2+1=25+1=26(số)

Tổng của dãy số 2;4;...;52 là:

\(\left(52+2\right)\cdot\frac{26}{2}=54\cdot13=702\)

(2+x)+(4+x)+...+(52+x)=780

=>26x+702=780

=>26x=78

=>x=3

p: \(70=2\cdot5\cdot7;80=2^4\cdot5\)

=>ƯCLN(70;80)=\(2\cdot5=10\)

70⋮x; 80⋮x

=>x∈ƯC(70;80)

=>x∈Ư(10)

mà x>8

nên x=10

q: \(12=2^2\cdot3;25=5^2;30=2\cdot3\cdot5\)

=>BCNN(12;25;30)=\(2^2\cdot3\cdot5^2=300\)

x⋮12; x⋮25; x⋮30

=>x∈BC(12;25;30)

=>x∈B(300)

mà 0<x<500

nên x=300