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\(\text{ĐKXĐ: }x+2\ne0\Leftrightarrow x\ne-2\)
\(x^2+\frac{4x^2}{\left(x+2\right)^2}=5\Leftrightarrow\frac{x^2\left(x+2\right)^2}{\left(x+2\right)^2}+\frac{4x^2}{\left(x+2\right)^2}=\frac{5\left(x+2\right)^2}{\left(x+2\right)^2}\)
\(\Leftrightarrow x^2\left(x+2\right)^2+4x^2=5\left(x+2\right)^2\)
<=>x2.(x2+4x+4)+4x2=5.(x2+4x+4)
<=>x4+4x3+4x2+4x2=5x2+20x+20
<=>x4+4x3+3x2-20x-20=0
<=>x4-2x3+6x3-12x2+15x2-30x+10x+20
<=>x3.(x-2)+6x2.(x-2)+15x.(x-2)+10.(x-2)=0
<=>(x-2)(x3+6x2+15x+10)=0
<=>(x-2)(x3+x2+5x2+5x+10x+10)=0
<=>(x-2).[x2(x+1)+5x.(x+1)+10.(x+1)]=0
<=>(x-2)(x+1)(x2+5x+10)=0
<=>x=2 hoặc x=-1 (vì x2+5x+10 = (x+5/2)2+15/4 >0)
Vậy S={-1;2}
\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)
\(\Leftrightarrow\frac{6}{x^2+2}-1+\frac{12}{x^2+8}-1=1-\frac{7}{x^2+3}\)
\(\Leftrightarrow\frac{6}{x^2+2}-\frac{x^2+2}{x^2+2}+\frac{12}{x^2+8}-\frac{x^2+8}{x^2+8}=\frac{x^2+3}{x^2+3}-\frac{7}{x^2+3}\)
\(\Leftrightarrow\frac{-x^2+4}{x^2+2}+\frac{-x^2+4}{x^2+8}=\frac{x^2-4}{x^2+3}\)
\(\Leftrightarrow\frac{-x^2+4}{x^2+2}+\frac{-x^2+4}{x^2+8}+\frac{-x^2+4}{x^2+3}=0\)
\(\Leftrightarrow\left(-x^2+4\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\right)=0\)
\(\Leftrightarrow-x^2+4=0\left(\text{vì : }\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\ne0\right)\)
<=>(2-x)(2+x)=0
<=>x=2 hoặc x=-2
Vậy S={-2;2}
\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10\)
\(\Leftrightarrow\frac{148-x}{25}-1+\frac{169-x}{23}-2+\frac{186-x}{21}-3+\frac{199-x}{19}-4=0\)
\(\Leftrightarrow\frac{148-x}{25}-\frac{25}{25}+\frac{169-x}{23}-\frac{46}{23}+\frac{186-x}{21}-\frac{63}{21}+\frac{199-x}{19}-\frac{76}{19}=0\)
\(\Leftrightarrow\frac{123-x}{25}+\frac{123-x}{23}+\frac{123-x}{21}+\frac{123-x}{19}=0\)
\(\Leftrightarrow\left(123-x\right).\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)
\(\Leftrightarrow123-x=0\left(\text{vì }\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\ne0\right)\)
<=>x=123
Vậy S={123}
Answer:
Câu 1:
\(\left(5x-x-\frac{1}{2}\right)2x\)
\(=\left(4x-\frac{1}{2}\right)2x\)
\(=4x.2x-\frac{1}{2}.2x\)
\(=8x^2-x\)
\(\left(x^3+4x^2+3x+12\right)\left(x+4\right)\)
\(=x\left(x^3+4x^2+3x+12\right)+4\left(x^3+4x^2+3x+12\right)\)
\(=x^4+4x^3+3x^2+12x+4x^3+16x^2+12x+48\)
\(=x^4+\left(4x^3+4x^3\right)+\left(3x^2+16x^2\right)+\left(12x+12x\right)+48\)
\(=x^4+8x^3+19x^2+24x+48\)
Ta thay \(x=99\) vào phân thức \(\frac{x^2+1}{x-1}\): \(\frac{\left(99\right)^2+1}{99-1}=\frac{9802}{98}=\frac{4901}{49}\)
Ta thay \(x=4\) vào phân thức \(\frac{x^2-x}{2\left(x-1\right)}\) : \(\frac{4^2-4}{2.\left(4-1\right)}=\frac{12}{6}=2\)
\(\left(x+y\right)^2-\left(x-y\right)^2\)
\(= (x²+2xy+y²)-(x²-2xy+y²)\)
\(= x²+2xy+y²-x²+2xy-y²\)
\(= 4xy\)
\(4x^2+4x+1=\left(2x+1\right)^2=\left(2.2+1\right)^2=25\)
Câu 2:
\(x^2+x=0\)
\(\Rightarrow x\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
\(x^2.\left(x-1\right)+4-4x=0\)
\(\Rightarrow x^2.\left(x-1\right)+4\left(1-x\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x^2-4\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\)
Trường hợp 1: \(x-1=0\Rightarrow x=1\)
Trường hợp 2: \(x-2=0\Rightarrow x=2\)
Trường hợp 3: \(x+2=0\Rightarrow x=-2\)
Câu 3: Bạn xem lại đề bài nhé.