ABC=2

Đề kiểm tra của mình nè

Câu 5:

Ta có:

\(\frac{1}{a^2+1}+\frac{1}{b^2+1}=\frac{2}{1+ab}\left(ĐKXĐ:a,b\in R\right)\)

\(\Leftrightarrow\frac{1}{a^2+1}+\frac{1}{b^2+1}-\frac{2}{1+ab}=0\)

\(\Leftrightarrow\left(\frac{1}{a^2+1}-\frac{1}{1+ab}\right)+\left(\frac{1}{b^2+1}-\frac{1}{1+ab}\right)=0\)

\(\Leftrightarrow\left[\frac{1+ab}{\left(a^2+1\right)\left(1+ab\right)}-\frac{a^2+1}{\left(a^2+1\right)\left(1+ab\right)}\right]\)\(+\left[\frac{1+ab}{\left(b^2+1\right)\left(1+ab\right)}-\frac{b^2+1}{\left(1+ab\right)\left(b^2+1\right)}\right]\)\(=0\)

\(\Leftrightarrow\frac{1+ab-a^2-1}{\left(a^2+1\right)\left(1+ab\right)}+\frac{1+ab-b^2-1}{\left(b^2+1\right)\left(1+ab\right)}=0\)

\(\Leftrightarrow\frac{ab-a^2}{\left(a^2+1\right)\left(1+ab\right)}+\frac{ab-b^2}{\left(b^2+1\right)\left(1+ab\right)}=0\)

\(\Leftrightarrow\frac{\left(ab-a^2\right)\left(b^2+1\right)}{\left(a^2+1\right)\left(1+ab\right)\left(b^2+1\right)}+\frac{\left(ab-b^2\right)\left(a^2+1\right)}{\left(b^2+1\right)\left(1+ab\right)\left(a^2+1\right)}=0\)

\(\Leftrightarrow\frac{\left(ab-a^2\right)\left(b^2+1\right)+\left(ab-b^2\right)\left(a^2+1\right)}{\left(a^2+1\right)\left(1+ab\right)\left(b^2+1\right)}=\frac{0}{\left(a^2+1\right)\left(1+ab\right)\left(b^2+1\right)}\)

\(\Rightarrow\left(ab-a^2\right)\left(b^2+1\right)+\left(ab-b^2\right)\left(a^2+1\right)=0\)

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

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

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

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

\(\Leftrightarrow ab\left(a-b\right)^2-\left(a-b\right)^2=0\)

\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2=0\)

\(\Leftrightarrow\orbr{\begin{cases}a-b=0\\ab-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}a=b\\ab=1\end{cases}}\Leftrightarrow a=b=\pm1\)

Lại có:

\(M=\frac{1}{a^{2021}+1}+\frac{1}{b^{2021}+1}\left(ĐK:a\ne-1;b\ne-1\right)\)

Mà ta có được \(a=b=\pm1\)nên thay \(a=b=1\)vào biểu thức M, ta được:

\(M=\frac{1}{1^{2021}+1}+\frac{1}{1^{2021}+1}=\frac{1}{1+1}+\frac{1}{1+1}=\frac{1}{2}+\frac{1}{2}=1\)

Vậy \(M=1\).

a, Xét tam giác ADB và tam giác BDC ta có : 

^ADB = ^BCD ( gt )

^ABD = ^BDC ( so le trong )

Vậy tam giác ADB ~ tam giác BDC ( g.g )

\(x^3-4x+1=\left(x-1\right)^2\)

\(\Leftrightarrow x^3-4x+1=x^2-2x+1\)

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

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

\(\Leftrightarrow\left(x\pm\sqrt{2}\right)\left(x-1\right)=0\Leftrightarrow x=\pm\sqrt{2};1\)

Vậy tập nghiệm của phương trình là S = { \(\pm\sqrt{2}\);1 } 

x³ - 4x + 1 = ( x - 1 )²

<=> x³ - x² - 2x = 0

<=> x( x² - x - 2 ) = 0

<=> x( x² - 2x + x - 2 ) = 0

<=> x[ x( x - 2 ) + ( x - 2 ) ] = 0

<=> x( x - 2 )( x + 1 ) = 0

<=> x = 0 hoặc x = 2 hoặc x = -1

Vậy pt có tập nghiệm S = { 0 ; 2 ; -1 }