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

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

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

c: \(\left|x^2+2x\right|+\left|y^2-9\right|=0\)

mà \(\left\{{}\begin{matrix}\left|x^2+2x\right|>=0\forall x\\\left|y^2-9\right|>=0\forall y\end{matrix}\right.\)

nên \(\left\{{}\begin{matrix}x^2+2x=0\\y^2-9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\left(x+2\right)=0\\\left(y-3\right)\left(y+3\right)=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x\in\left\{0;-2\right\}\\y\in\left\{3;-3\right\}\end{matrix}\right.\)

d: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=120\)

=>\(2^x\left(1+2+2^2+2^3\right)=120\)

=>\(2^x\cdot15=120\)

=>\(2^x=8\)

=>x=3

e: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

=>\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)

=>\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)

=>\(\left[{}\begin{matrix}x-7=0\\x-7=1\\x-7=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\\x=6\end{matrix}\right.\)

a: \(8x\left(2023+x\right)-8x\left(x+2024\right)=56\)

=>\(8x\left(x+2023-x-2024\right)=56\)

=>-8x=56

=>\(x=\frac{56}{-8}=-7\) (nhận)

b: \(5-2025x=9-2026\left(x-1\right)\)

=>-2025x+5=9-2026x+2026

=>-2025x+5=-2026x+2035

=>x=2035-5=2030(nhận)

c: \(\left(-12\right)^2\cdot x=56+10\cdot13x\)

=>144x=56+130x

=>14x=56

=>x=4(nhận)

d: -(x-32+11)=(21-33-x+7)

=>-(x-21)=(-x-5)

=>-x+21=-x-5

=>21=-5(vô lý)

=>x∈∅

e: \(-2\left(x+6\right)+6\left(x+10\right)=8\)

=>-2x-12+6x+60=8

=>4x+52=8

=>4x=8-52=-44

=>\(x=-\frac{44}{4}=-11\) (nhận)

1.     Giải:

Do \(5x+13B\in\left(2x+1\right)\Rightarrow5x+13⋮2x+1.\)

 \(\Rightarrow2\left(5x+13\right)⋮2x+1\Rightarrow10x+26⋮2x+1.\)

 \(\Rightarrow5\left(2x+1\right)+21⋮2x+1.\)

Do 5(2x+1)⋮2x+1⇒ Ta cần 21⋮2x+1.

⇒ 2x+1 ϵ B(21)=\(\left\{1;3;7;21\right\}.\)

Ta có bảng:

Vậy xϵ\(\left\{0;1;3;10\right\}.\)

2. Giải:

Do (2x-18).(3x+12)=0.

⇒ 2x-18=0             hoặc             3x+12=0.

⇒ 2x     =18                               3x       =-12.

⇒   x     =9                                   x       =-4.

Vậy xϵ\(\left\{-4;9\right\}.\)

3. S= 1-2-3+4+5-6-7+8+...+2021-2022-2023+2024+2025.

S= (1-2-3+4)+(5-6-7+8)+...+(2021-2022-2023+2024)+2025 Có 506 cặp.

S= 0 + 0 + ... + 0 + 2025.

⇒S= 2025.

a:

Sửa đề: \(S=1-3+5-7+...+2021-2023+2025\)

Từ 1 đến 2025 sẽ có:

\(\dfrac{2025-1}{2}+1=\dfrac{2024}{2}+1=1013\left(số\right)\)

Ta có: 1-3=5-7=...=2021-2023=-2

=>Sẽ có \(\dfrac{1013-1}{2}=\dfrac{1012}{2}=506\) cặp có tổng là -2 trong dãy số này

=>\(S=506\cdot\left(-2\right)+2025=2025-1012=1013\)

b: \(S=1+2-3-4+5+6-7-8+...+2021+2022-2023-2024\)

Từ 1 đến 2024 là: \(\dfrac{\left(2024-1\right)}{1}+1=2024\left(số\right)\)

Ta có: 1+2-3-4=5+6-7-8=...=2021+2022-2023-2024=-4

=>Sẽ có \(\dfrac{2024}{4}=506\) cặp có tổng là -4 trong dãy số này

=>\(S=506\cdot\left(-4\right)=-2024\)

1-2+3-4+5-6+7-8+...+2023-2024

=(1−2)+(3−4)+(5−6)+(7−8)+....+(2023−2024)=(1−2)+(3−4)+(5−6)+(7−8)+....+(2023−2024)

=−1+(−1)+(−1)+(−1)+...+(−1)=−1+(−1)+(−1)+(−1)+...+(−1)

=−1.1012=−1.1012

=−1012=−1012

1-2+3-4+5-6+ ... +2023-2024

= (-1) + (-1) + ... + (-1) (1012 số)

= (-1).1012

= -1012

helps me

^-^

bài rất dễ hỉu =))))))))

Ta có: \(A=2^3\cdot5^3-6\cdot\left\lbrace423-\left\lbrack673-2^3\cdot\left(7^{18}:7^{16}+2024^0\right)\right\rbrack\right\rbrace\)

\(=10^3-6\cdot\left\lbrace423-\left\lbrack673-8\cdot\left(7^2+1\right)\right\rbrack\right\rbrace\)

\(=1000-6\cdot\left\lbrace423-\left\lbrack673-8\cdot50\right\rbrack\right\rbrace\)

\(=1000-6\cdot\left\lbrace423-\left\lbrack673-400\right\rbrack\right\rbrace\)

\(=1000-6\cdot\left\lbrace423-273\right\rbrace\)

\(=1000-6\cdot150=1000-900=100\)

P=[(1-2)+(-3+4)+(5-6)+(-7+8)+...+(993-994)+(-995+996)]+997

P=[(-1)+1+(-1)+1+...+(-1)+1+(-1)+1]+997

P= 0 +0 +...+ 0 +997

P=997