lm trên symbolab (symbolab.com)

Bài làm:

Đặt \(x+4=y\)

\(Pt\Leftrightarrow\left(y-1\right)^4+\left(y+1\right)^4=2\)

\(\Leftrightarrow y^4-4y^3+6y^2-4y+1+y^4+4y^3+6y^2+4y+1-2=0\)

\(\Leftrightarrow2y^4+12y^2=0\)

\(\Leftrightarrow2y^2\left(y^2+6\right)=0\)

Mà \(y^2+6\ge6>0\left(\forall y\right)\)

\(\Rightarrow y^2=0\Leftrightarrow y=0\Leftrightarrow x+4=0\Rightarrow x=-4\)

Vậy \(x=-4\)

Đặt: x -1 = a; x + 2 = b

=> 2x + 1 = a + b

=> Ta có pt mới: 

\(a^3+b^3=\left(a+b\right)^3\)

<=> \(3ab\left(a+b\right)=0\)

<=> \(\orbr{\begin{cases}a=0\\b=0\end{cases}}\)hoặc a + b = 0

=> x-1=0 hoặc x+2=0 hoặc 2x+1=0

<=> x=1 hoặc x=-2 hoặc x=-1/2.

<=> [(x-1)+(x+2)].[(x-1)² - (x-1).(x+2) + (x+2)² ] = (2x+1)² 

<=> (2x+1).[x² -2x+1-(x² -x-2)+x² +4x+4] = (2x+1)³ 

<=> x² -2x+1-x² +x+2+x² +4x+4 = 4x² +4x+1 (x khác -1/2)

<=> 3x² +x-6=0 đến đây là PT bậc 2 rồi bạn tự làm nốt

Ta có : \(\hept{\begin{cases}A=1999.2001\\B=2000^2\end{cases}}\)

\(< =>\hept{\begin{cases}A=1999.2000+1999\\B=2000\cdot2000\end{cases}}\)

\(< =>\hept{\begin{cases}A=1999.2000+2000+1\\B=1999.2000+2000\end{cases}}\)

\(< =>\hept{\begin{cases}A=2000.2000+1\\B=2000.2000\end{cases}}\)

\(< =>A>B\)

a. Ta có : \(A=1999.2021=\left(2000-1\right)\left(2000+1\right)=2020^2-1< 2020\)

\(\Rightarrow A< B\)

b. Ta có : \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

...

\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}\)

\(\Rightarrow A>B\)

c,d tương tự

Bài làm:

Đặt \(\hept{\begin{cases}x-1=a\\x+2=b\end{cases}}\Rightarrow a+b=2x+1\)

\(Pt\Leftrightarrow a^3+b^3=\left(a+b\right)^3\)

\(\Leftrightarrow a^3+3a^2b+3ab^2+b^3-a^3-b^3=0\)

\(\Leftrightarrow3ab\left(a+b\right)=0\)

\(\Leftrightarrow3\left(x-1\right)\left(x+2\right)\left(2x+1\right)=0\)

Đến đây giải PT tích ra ta được: \(x\in\left\{-2;-\frac{1}{2};1\right\}\)

a) 4x³ - 14x² + 6x

= 2x( 2x² - 7x + 3 )

= 2x( 2x² - 6x - x + 3 )

= 2x[ 2x( x - 3 ) - ( x - 3 ) ]

= 2x( 2x - 1 )( x - 3 )

b) x⁴ + 4

= x⁴ + 4 + 4x² - 4x²

= ( x⁴ + 4x² + 4 ) - 4x²

= ( x² + 2 )² - ( 2x )²

= ( x² + 2 + 2x )( x² + 2 - 2x )  

a) 

\(=2x\left(2x^2-7x+3\right)\)

\(=2x\left(x-3\right)\left(2x-1\right)\)

b) 

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

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

a) \(A⋮B\Leftrightarrow n\ge2\)\(\left(n\in Z\right)\)

b) \(A⋮B\Leftrightarrow2n\ge2\Leftrightarrow n\ge1\)\(\left(n\in Z\right)\)

Bg

x³ - 3x - x(x² + 5x + 2) = 0

(x² - 3).x - x(x² + 5x + 2) = 0

x.[x² - 3 - (x² + 5x + 2)] = 0

=> x = 0 hoặc x² - 3 - (x² + 5x + 2) = 0

Xét x² - 3 - (x² + 5x + 2) = 0:

=> x² - 3 - x² - 5x - 2 = 0

=> x² - x² - (3 + 2) - 5x = 0

=> -5 - 5x = 0

=> 5x = -5

=> x = -5 : 5

=> x = -1

Vậy x = 0 hoặc x = -1

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

\(\Leftrightarrow x^3-3x-x^3-5x^2-2x=0\)

\(\Leftrightarrow-5x-5x^2=0\)

\(\Leftrightarrow-5x\left(1+x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}-5x=0\\1+x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

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

\(P'=\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\)    (​     \(P=\frac{1}{P'}\))

\(P'=\frac{x+2-\left(x-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(P'=\frac{-x-\sqrt{x}+2}{\left(\sqrt{x}-1\right) \left(x+\sqrt{x}+1\right)}=\frac{\left(1-\sqrt{x}\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=-\frac{\sqrt{x}+2}{x+\sqrt{x}+1}\)

=> \(P=-\frac{x+\sqrt{x}+1}{\sqrt{x}+2}\)

Hermit ĐKXĐ đâu bạn eiii ??

pt <=> \(x^2+21x+90=136\)

<=> \(x^2+21x-46=0\)

<=> \(x^2-2x+23x-46=0\)

<=> \(\left(x-2\right)\left(x+23\right)=0\)

<=> \(\orbr{\begin{cases}x=2\\x=-23\end{cases}}\)

Vậy tập hợp nghiệm của pt là S={2;-23}.

\(\left(15+x\right)\left(6+x\right)=136\)

\(\Leftrightarrow90+15x+6x+x^2=136\)

\(\Leftrightarrow-46+21x+x^2=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+23\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-23\end{cases}}\)