1, ( x - 3 )¹⁰⁰ - 1 = 0 => ( x - 3 )¹⁰⁰ = 0 + 1 = 1

Mà 1 = 1¹⁰⁰ => x - 3 = 1 => x = 1 + 3 = 4 hoặc 1 = (-1)¹⁰⁰ => x - 3 = -1 => x = -1 + 3 = 2

2, ( 9 - 5x )²⁰¹⁹ + 1 = 0 => ( 9 - 5x )²⁰¹⁹ = 0 - 1 = -1

Mà -1 = (-1)²⁰¹⁹ => 9 - 5x = -1 => 5x = 9 - ( -1 ) = 10 => x = 10 : 5 = 2

3, ( 4x - 3 )⁵ = ( 2 - x )⁵ => 4x - 3 = 2 - x

=> 4x + x = 3 + 2 => 5x = 5 => x = 5 : 5 = 1

4, ( 2x - 9 )² = ( 5x - 6 )² => 2x - 9 = 5x - 6 ... ( tự làm )

5, ( 11 - 4x )⁶ - ( 2 - 5x )⁶ = 0 => ( 11- 4x )⁶ = ( 2 - 5x )⁶

=> 11 - 4x = 2 - 5x _ Đến đây làm tương tự 2 câu trên

6, ( x - 9 )⁹ = ( x - 9 )⁷ mà cơ số bằng nhau ( = x - 9 )

=> x - 9 = 1 hoặc -1 vì 1⁹ = 1⁷ và ( -1 )⁹ = ( -1 )⁷

TH1: x - 9 = 1 => x = 1 + 9 = 10

TH2: x - 9 = -1 => x = -1 + 9 = 8

7, 8, 9 tương tự 6 ( kết quả của cơ số đều = 1 hoặc -1 )

\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)

\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)

\(A=1-\frac{1}{2020}\)

\(A=\frac{2019}{2020}\)

\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)

\(2B=\frac{2}{1.3}+\frac{2}{3.5}=\frac{2}{5.7}+...+\frac{2}{2017.2019}\)

\(2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}=\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)

\(2B=1-\frac{1}{2019}\)

\(2B=\frac{2018}{2019}\)

\(B=\frac{2018}{2019}:2=\frac{1009}{2019}\)

\(C=3^0+3^1+3^2+...+3^{100}\)

\(3C=3^1+3^2+3^3+...+3^{101}\)

\(3C-C=\left(3^1+3^2+3^3+...+3^{101}\right)-\left(1+3^1+3^2+...+3^{100}\right)\)

\(2C=3^{101}-1\)

\(C=\frac{3^{101}-1}{2}\)

\(x^{2016}=x^{2017}\\ \Leftrightarrow x^{2016}-x^{2017}=0\\ \Leftrightarrow x^{2016}\left(1-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^{2016}=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy S = {0;1}

5^x=5^2019:(5^2013-100.5^2010)

=> 5^x=5²⁰¹⁹:(5²⁰¹⁰.(5³-100))

=>5^x=5²⁰¹⁹:(5²⁰¹⁰.(125-100))

=>5^x=5²⁰¹⁹:(5²⁰¹⁰.25)

=>5^x=5²⁰¹⁹:5²⁰¹²

^=>5^x=5⁷

=> x=7

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)

a,\(\text{Để }\left(4x+3\right)^3-\left(2x-5\right)^3=\left(2x+8\right)^3\) thì

\(3\left(4x+3\right)\left(2x-5\right)\left(2x+8\right)=0\)

\(\Leftrightarrow\left(4x+3\right)\left(2x-5\right)\left(2x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+3=0\\2x-5=0\\2x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{5}{2}\\x=-4\end{matrix}\right.\)

Vậy..

b,\(Để\left(3x+2016\right)^3+\left(3x-2019\right)^3=\left(6x-3\right)^3\) thì

\(3\left(3x+2016\right)\left(3x-2019\right)\left(6x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2016=0\\3x-2019=0\\6x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2016}{3}\\x=\dfrac{2019}{3}\\x=\dfrac{1}{2}\end{matrix}\right.\)

Vậy...

Giúp mình nhé các bạn mình đang cần gấp lắm

d, \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)

\(\Leftrightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)

\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)

\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)

\(\Leftrightarrow x+10=0\) (Vì \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\) ≠ 0)

\(\Leftrightarrow x=-10\)

Vậy x = -10 là nghiệm của phương trình.

e, Đề sai.

Sửa đề: \(\frac{x}{2016}+\frac{x+1}{2017}+\frac{x+2}{2018}+\frac{x+3}{2019}+\frac{x+4}{2020}=5\)

\(\Leftrightarrow\frac{x}{2016}-1+\frac{x+1}{2017}-1+\frac{x+2}{2018}-1+\frac{x+3}{2019}-1+\frac{x+4}{2020}-1=0\)

\(\Leftrightarrow\frac{x-2016}{2016}+\frac{x-2016}{2017}+\frac{x-2016}{2018}+\frac{x-2016}{2019}+\frac{x-2016}{2020}=0\)

\(\Leftrightarrow\left(x-2016\right)\left(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}+\frac{1}{2020}\right)=0\)

\(\Leftrightarrow x-2016=0\) (Vì \(\frac{1}{2016}+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}+\frac{1}{2020}\ne0\)

\(\Leftrightarrow x=2016\)

Vậy x = 2016 là nghiệm của phương trình.

Ta có: \(\frac{x-5}{2016}+\frac{x-4}{2017}=\frac{x-3}{2018}+\frac{x-2}{2019}\)

\(\Leftrightarrow\frac{x-5}{2016}-1+\frac{x-4}{2017}-1=\frac{x-3}{2018}-1+\frac{x-2}{2019}-1\)

\(\Leftrightarrow\frac{x-2021}{2016}+\frac{x-2021}{2017}=\frac{x-2021}{2018}+\frac{x-2021}{2019}\)

\(\Leftrightarrow\frac{x-2021}{2016}+\frac{x-2021}{2017}-\frac{x-2021}{2018}-\frac{x-2021}{2019}=0\)

\(\Leftrightarrow\left(x-2021\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\)

mà \(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\ne0\)

nên x-2021=0

hay x=2021

Vậy: x=2021

Ta có:

\(\frac{x-5}{2016}+\frac{x-4}{2017}=\frac{x-3}{2018}+\frac{x-2}{2019}\)

\(\Leftrightarrow\left(\frac{x-5}{2016}-1\right)+\left(\frac{x-4}{2017}-1\right)=\left(\frac{x-3}{2018}-1\right)+\left(\frac{x-2}{2019}-1\right)\)

\(\Leftrightarrow\frac{x-2021}{2016}+\frac{x-2021}{2017}=\frac{x-2021}{2018}+\frac{x-2021}{2019}\)

\(\Leftrightarrow\frac{x-2021}{2016}+\frac{x-2021}{2017}-\frac{x-2021}{2018}-\frac{x-2021}{2019}=0\)

\(\Leftrightarrow\left(x-2021\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\)

Vì \(\frac{1}{2016}>\frac{1}{2018}\) và \(\frac{1}{2017}>\frac{1}{2019}\) nên \(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}>0\)

\(\Rightarrow x-2021=0\)

\(\Rightarrow x=2021\)

Vậy phương trình có nghiệm duy nhất là \(x=2021\)