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a) \(2\left(x-1\right)^2+\left(x+3\right)^2=3\left(x-2\right)\left(x+1\right)\)
\(\Leftrightarrow2x^2-4x+2+x^2+6x+9=3x^2-3x-6\)
\(\Leftrightarrow5x=-17\)
\(\Rightarrow x=-\frac{17}{5}\)
b) \(\left(x+2\right)^2-2\left(x-3\right)=\left(x+1\right)^2\)
\(\Leftrightarrow x^2+4x+4-2x+6=x^2+2x+1\)
\(\Leftrightarrow10=1\)
=> vô nghiệm
c) \(\left(x-1\right)^2+\left(x-2\right)^2=2\left(x+4\right)^2-\left(22x+27\right)\)
\(\Leftrightarrow x^2-2x+1+x^2-4x+4=2x^2+8x+8-22x-27\)
\(\Leftrightarrow8x=-24\)
\(\Rightarrow x=-3\)
a) 2( x - 1 )2 + ( x + 3 )2 = 3( x - 2 )( x + 1 )
<=> 2( x2 - 2x + 1 ) + x2 + 6x + 9 = 3( x2 - x - 2 )
<=> 2x2 - 4x + 2 + x2 + 6x + 9 = 3x2 - 3x - 6
<=> 2x2 - 4x + x2 + 6x - 3x2 + 3x = -6 - 2 - 9
<=> 5x = -17
<=> x = -17/5
b) ( x + 2 )2 - 2( x - 3 ) = ( x + 1 )2
<=> x2 + 4x + 4 - 2x + 6 = x2 + 2x + 1
<=> x2 + 4x - 2x - x2 - 2x = 1 - 4 - 6
<=> 0x = -9 ( vô lí )
Vậy phương trình vô nghiệm
c) ( x - 1 )2 + ( x - 2 )2 = 2( x + 4 )2 - ( 22x + 27 )
<=> x2 - 2x + 1 + x2 - 4x + 4 = 2( x2 + 8x + 16 ) - 22x - 27
<=> 2x2 - 6x + 5 = 2x2 + 16x + 32 - 22x - 27
<=> 2x2 - 6x - 2x2 - 16x + 22x = 32 - 27 - 5
<=> 0x = 0 ( đúng ∀ x ∈ R )
Vậy phương trình nghiệm đúng ∀ x ∈ R
A) Với \(x>y>0\),ta có: \(x^2+y^2< x^2+y^2+2xy=\left(x+y\right)^2\Rightarrow\frac{1}{x^2+y^2}>\frac{1}{\left(x+y\right)^2}\)
Xét: \(\frac{x^2-y^2}{x^2+y^2}>\frac{x^2-y^2}{\left(x+y\right)^2}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}=\frac{x-y}{x+y}\)--->ĐPCM
B) \(3^{16}+1=\left(3^{16}-1\right)+2=\left(3^8+1\right)\left(3^8-1\right)+2\)
\(=\left(3^8+1\right)\left(3^4+1\right)\left(3^4-1\right)+2\)
\(=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3^2-1\right)+2\)
\(=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\left(3-1\right)+2\)
\(>\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\)--->ĐPCM
\(x^3+2x^2+3x=0\)\(\Leftrightarrow x.\frac{x^3+2x^2+3x}{x}=0\)
\(\Leftrightarrow x\left(x^2+2x+3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2+2x+3=0\end{cases}}\)
Ta sẽ c/m \(x^2+2x+3=0\) vô nghiệm.Thật vậy:
\(x^2+2x+3=\left(x+1\right)^2+2\ge2\forall x\)
Từ đó suy ra \(x^2+2x+3=0\) vô nghiệm.
Vậy : x = 0
\(\left(x+2\right)\left(2x-1\right)+1=4x^2\)
\(2x^2-x+4x-2+1=4x^2\)
\(\Rightarrow2x^2-3x+1=0\)
\(2x\left(x-1\right)-\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}}\)
Vậy \(\orbr{\begin{cases}x=1\\x=\frac{1}{2}\end{cases}}\)
ý còn lại tham khảo bài tth
Ta có: \(\left(x+x^2\right)^{2+1}=0\)
\(\Leftrightarrow\left[x\left(x+1\right)\right]^3=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)