\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)

\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=4\)

\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}=\frac{2}{xy}-\frac{1}{z^2}\)

\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}-\frac{2}{xy}+\frac{1}{z^2}=0\)

\(\Leftrightarrow\left(\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\frac{1}{x}+\frac{1}{z}=0\\\frac{1}{y}+\frac{1}{z}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}\frac{1}{x}=\frac{1}{-z}\\\frac{1}{y}=\frac{1}{-z}\end{cases}\Leftrightarrow}\frac{1}{x}=\frac{1}{y}=\frac{1}{-z}\)

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)

\(\Leftrightarrow\frac{1}{-z}+\frac{1}{-z}+\frac{1}{z}=2\)

\(\Leftrightarrow z=\frac{-1}{2}\)

\(x=y=\frac{1}{2}\)

\(\Rightarrow C=\left(x+2y+z\right)^{2021}=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2021}=1^{2021}=1\)

Ta có:\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\Rightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=4\)

\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2=\frac{2}{xy}-\frac{1}{z^2}\)

\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}-\frac{2}{xy}+\frac{1}{z^2}=0\)

\(\Leftrightarrow\left(\frac{1}{x^2}+\frac{2}{xz}+\frac{1}{z^2}\right)+\left(\frac{1}{y^2}+\frac{2}{yz}+\frac{1}{z^2}\right)=0\)

\(\Leftrightarrow\left(\frac{1}{x}+\frac{1}{z}\right)^2+\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\left(\frac{1}{x}+\frac{1}{z}\right)^2=0\\\left(\frac{1}{y}+\frac{1}{z}\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{1}{x}=-\frac{1}{z}\\\frac{1}{y}=-\frac{1}{z}\end{cases}}}\)

\(\Leftrightarrow x=y=-z\)

Thay vào \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2\)ta được :

\(x=y=\frac{1}{2};z=-\frac{1}{2}\)

\(\Rightarrow P=\left(\frac{1}{2}+2.\frac{1}{2}-\frac{1}{2}\right)^{2021}=1^{2020}=1\)

Pt <=> y^3 =3x + x^3

Vì 3x^2 + 1 > 0 mọi x nên ta có:

(X^3 +3x ) - (3x^2 + 1) < x^3 + 3x < x^3 + 3x + (3x^2 + 1)

<=> (x-1)^3 < y^3 < (x + 1)^3

=> y^3 =x^3

Pt <=>x^3 =x^3 + 3x

<=> x = 0

=> y= 0 Vậy ngiệm của pt là (0,0)

\(y^3-x^3=3x\)

\(\Leftrightarrow x^3+3x=y^3\)

Vì \(3x^2+1>0\forall x\)ta có:

\(\left(x^3+3x\right)-\left(3x^2+1\right)< x^3+3x+\left(3x^2+1\right)\)

\(\Leftrightarrow\left(x-1\right)^3< y^3< \left(x+1\right)^3\)

\(\Rightarrow x^3=y^3\)

Ta có:\(y^3-x^3=3x\)

\(\Leftrightarrow x^3+3x=x^3\)

\(\Rightarrow x=0\)

\(\Rightarrow x=y=0\)

Cre:mạng

a) pt <=> ( x - 1 )³ + x²( x - 1 ) = 0

<=> ( x - 1 )[ ( x - 1 )² + x² ] = 0

<=> x = 1

Vậy pt có nghiệm x = 1

b) x² + x - 12 = 0

<=> x² - 3x + 4x - 12 = 0

<=> x( x - 3 ) + 4( x - 3 ) = 0

<=> ( x - 3 )( x + 4 ) = 0

<=> x = 3 hoặc x = -4

Vậy S = { 3 ; -4 }

c) x + x⁴ = 0

<=> x( x³ + 1 ) = 0

<=> x( x + 1 )( x² - x + 1 ) = 0

<=> x = 0 hoặc x = -1

Vậy S = { 0 ; -1 }

a,\(x^3-3x^2+3x-1+x\left(x^2-x\right)=0\)

\(\Leftrightarrow\left(x^3-3x^2+3x-1\right)+x\left(x^2-x\right)=0\)

\(\Leftrightarrow\left(x-1\right)^3+x^2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x-1\right)^2+x^2\right]=0\)

\(\Leftrightarrow x=1\)

\(\orbr{\begin{cases}\\\end{cases}}\)

ĐKXĐ : x khác 3

\(\frac{x+4}{x-3}=1\Rightarrow x+4=x-3\Leftrightarrow x-x=-3-4\Leftrightarrow0=-7\)( vô lí )

Vậy pt vô nghiệm

Trả lời:

\(\frac{x+4}{x-3}=1\)\(\left(đkxđ:x\ne3\right)\)

\(\Leftrightarrow x+4=x-3\)

\(\Leftrightarrow x-x=-3-4\)

\(\Leftrightarrow0x=-7\)(vô lí)

Vậy \(S=\varnothing\)

\(\left(n^2+n-1\right)^2-1=\left(n^2+n-2\right)\left(n^2+n\right)\)

\(=\left(n-1\right)\left(n+2\right)n\left(n+1\right)⋮2,3,4\)

Mà (2;3;4)=1

=> \(\left(n-1\right)\left(n+2\right)n\left(n+1\right)⋮24\)

Ta có: \(\left(n^2+n-1\right)^2-1=\left[n^2+\left(n-1\right)\right]^2-1\)

                                                   \(=n^4+2.n^2.\left(n-1\right)+\left(n-1\right)^2-1\)

                                                   \(=n^4+2n^3-n^2-2n\)

                                                   \(=n^3.\left(n +2\right)-n.\left(n+2\right)\)

                                                   \(=n.\left(n^2-1\right).\left(n+2\right)\)

                                                   \(=\left(n-1\right).n.\left(n+1\right).\left(n+2\right)\)

Ta nhận thấy\(\left(n-1\right).n.\left(n+1\right).\left(n+2\right)\)là tích của 4 số nguyên liên tiếp 

mà tích của 4 số nguyên liên tiếp thì luôn chia hết cho 24

\(\Rightarrow\)\(\left(n-1\right).n.\left(n+1\right).\left(n+2\right)⋮24\forall x\inℤ\)

hay \(\left(n^2+n-1\right)^2-1⋮24\forall x\inℤ\)

\(A=\left(\frac{x^3-1}{x^2-x}+\frac{x^2-4}{x^2-2x}-\frac{2-x}{x}\right)\div\frac{x+1}{x}\)

a) ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne2\end{cases}}\)

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

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

\(=\frac{x^2+3x+1}{x}\times\frac{x}{x+1}=\frac{x^2+3x+1}{x+1}\)

b) x³ - 4x² + 3x = 0

<=> x( x² - 4x + 3 ) = 0

<=> x( x - 1 )( x - 3 ) = 0

<=> x = 0 (ktm) hoặc x = 1(tm) hoặc x = 3(tm)

Bạn tự thế các giá trị tm nhé ;)

b) Ta có: \(x^3-4x^2+3x=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x-3\right)=0\)

<=> x=0 ( loại) hoặc x=1 (loại) hoặc x=3 ( thỏa mãn)

Thay x=3 vào A ta có:

\(A=\frac{3^2+3.3+1}{3+1}=\frac{19}{4}\)

\(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)

\(\Leftrightarrow\frac{9}{8}-\frac{3}{8}x+\frac{10}{3}-\frac{2}{3}x=\frac{1}{2}-\frac{1}{2}x-2\)

\(\Leftrightarrow-\frac{3}{8}x-\frac{2}{3}x+\frac{1}{2}x=\frac{1}{2}-2-\frac{9}{8}-\frac{10}{3}\)

\(\Leftrightarrow-\frac{13}{24}x=-\frac{143}{24}\)

\(\Leftrightarrow x=11\)

Trả lời:

\(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)

\(\Leftrightarrow\frac{9\left(3-x\right)}{24}+\frac{16\left(5-x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)

\(\Leftrightarrow27-9x+80-16x=12-12x-48\)

\(\Leftrightarrow107-25x=-36-12x\)

\(\Leftrightarrow-25x+12x=-36-107\)

\(\Leftrightarrow-13x=-143\)

\(\Leftrightarrow x=11\)

Vậy \(S=\left\{11\right\}\)

-Dạng 1: Phương trình tích.

a) \(2x\left(x+1\right)=x^2-1\)\(\Leftrightarrow2x\left(x+1\right)=\left(x-1\right)\left(x+1\right)\)

\(\Leftrightarrow2x\left(x+1\right)-\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(2x-x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x+1\right)=0\)

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

\(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy phương trình có nghiệm duy nhất : x = -1

b) \(x^3+3x^2-2x-2=0\)\(\Leftrightarrow\left(x^3-x^2\right)+\left(4x^2-4x\right)+\left(2x-2\right)=0\)

\(\Leftrightarrow x^2\left(x-1\right)+4x\left(x-1\right)+2\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^2+4x+2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[\left(x+2\right)^2-2\right]\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+2\right)^2-2=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\left(x+2\right)^2=2\\x=1\end{cases}}}\)

Xét phương trình \(\left(x+2\right)^2=2\)

\(\Leftrightarrow\orbr{\begin{cases}x+2=\sqrt{2}\\x+2=-\sqrt{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\sqrt{2}-2\\x=-\sqrt{2}-2\end{cases}}}\)

Vậy phương trình có tập nghiệm : \(S=\left\{1;\pm\sqrt{2}-2\right\}\)

Dạng 2  ; Phương trình chứa ẩn ở mẫu.

\(\frac{3}{1-5x}+\frac{5}{3-5x}=\frac{x-27}{\left(5x-1\right)\left(5x-3\right)}\left(ĐKXĐ:x\ne\frac{1}{5};x\ne\frac{3}{5}\right)\)

\(\Leftrightarrow\frac{3}{1-5x}+\frac{5}{3-5x}=\frac{x-27}{\left(1-5x\right)\left(3-5x\right)}\)(phần này bạn nhớ đọc kĩ bên vế phải)

\(\Leftrightarrow\frac{3\left(3-5x\right)}{\left(1-5x\right)\left(3-5x\right)}+\frac{5\left(1-5x\right)}{\left(3-5x\right)\left(1-5x\right)}=\frac{x-27}{\left(1-5x\right)\left(3-5x\right)}\)

\(\Rightarrow3\left(3-5x\right)+5\left(1-5x\right)=x-27\)

\(\Leftrightarrow9-15x+5-25x=x-27\)

\(\Leftrightarrow14-40x=x-27\)

\(\Leftrightarrow-40x-x=-27-14\)

\(\Leftrightarrow-41x=-41\)

\(\Leftrightarrow x=1\)(thỏa mãn ĐKXĐ)

Vậy phương trình có nghiệm duy nhất : x = 1.

\(3^{2n}-9\)\(⋮72\)

\(3^{2n}-9=\left(3^2\right)^n-9\)\(=9^n-9⋮9\)\(\left(1\right)\)

Ta có: 

 \(9\)đồng dư với 1 (mod 8)

\(9^n\)đồng dư với 1 (mod 8)

\(9^n-9\)đồng dư với -8 (mod 8)

\(9^n-9\)đồng dư với 0 (mod 8)

\(9^n-9\)\(⋮8\)\(\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)ta suy ra:

\(3^{2n}-9\)\(⋮72\)