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a) \(\sqrt{4\left(a-3\right)^2}=\sqrt{2^2\left(a-3\right)^2}=2\sqrt{\left(a-3\right)^2}=2.\left|a-3\right|=2\left(a-3\right)=2a-6\) (Vì \(a\ge3\) )
b) \(\sqrt{9\left(b-2\right)^2}=\sqrt{3^2\left(b-2\right)^2}=3\sqrt{\left(b-2\right)^2}=3\left|b-2\right|=3\left(2-b\right)\)
\(=6-3b\) (vì b < 2 )
b) \(\sqrt{27.48\left(1-a\right)^2}=\sqrt{27.3.16.\left(1-a\right)^2}=\sqrt{81.16.\left(1-a\right)^2}\)
\(=\sqrt{9^2.4^2.\left(1-a\right)^2}=9.4\sqrt{\left(1-a\right)^2}=36.\left|1-a\right|=36\left(1-a\right)=36-36a\) (vì a > 1)
Ta có: \(x^2-5x+3=0\)
Áp dụng định lí viet ta có: \(\hept{\begin{cases}x_1+x_2=5\\x_1x_2=3\end{cases}}\)
a) \(A=x_1^2+x_2^2=\left(x_1+x_2\right)^2-2x_1x_2=5^2-2.3=19\)
b) \(B=x_1^3+x_2^3=\left(x_1+x_2\right)^3-3\left(x_1+x_2\right)x_1x_2=5^3-3.5.3=80\)
c) \(C=\left|x_1-x_2\right|\)>0
=> \(C^2=x_1^2+x_2^2-2x_1x_2=19-2.3=13\)
=> C = căn 13
d) \(D=x_2+\frac{1}{x_1}+x_1+\frac{1}{x_2}=\left(x_1+x_2\right)+\frac{x_1+x_2}{x_1x_2}=5+\frac{5}{3}=5\frac{5}{3}\)
e) \(E=\frac{1}{x_1+3}+\frac{1}{x_2+3}=\frac{\left(x_1+x_2\right)+6}{x_1x_2+3\left(x_1+x_2\right)+9}=\frac{5+6}{3+3.5+9}=\frac{11}{27}\)
g) \(G=\frac{x_1-3}{x_1^2}+\frac{x_2-3}{x_2^2}=\left(\frac{1}{x_1}+\frac{1}{x_2}\right)-3\left(\frac{1}{x_1^2}+\frac{1}{x_2^2}\right)\)
\(=\frac{x_1+x_2}{x_1x_2}-3\frac{x_1^2+x_2^2}{x_1^2.x_2^2}=\frac{5}{3}-3.\frac{19}{3^2}=-\frac{14}{3}\)
+) ĐK: x khác -5
\(x^2+\frac{25x^2}{\left(x+5\right)^2}=11\)
<=> \(x^2+\frac{25x^2}{\left(x+5\right)^2}-2.x\frac{5x}{\left(x+5\right)}+\frac{10x^2}{\left(x+5\right)}=11\)
<=> \(\left(x-\frac{5x}{x+5}\right)^2+\frac{10x^2}{x+5}=11\)
<=> \(\left(\frac{x^2}{x+5}\right)^2+\frac{10x^2}{x+5}-11=0\) ( đặt t = x^2/x+5 => có phương trình: t^2 + 10t - 11 = 0 => giải t => tìm x )
<=> \(\orbr{\begin{cases}\frac{x^2}{x+5}=1\\\frac{x^2}{x+5}=-11\end{cases}}\Leftrightarrow\orbr{\begin{cases}x^2-x-5=0\\x^2+11x+55=0\left(vn\right)\end{cases}}\Leftrightarrow x=\frac{1}{2}\pm\frac{\sqrt{21}}{2}\) ( thỏa mãn)
\(x^2+\frac{25x^2}{\left(x+5\right)^2}=11ĐK:x\ne-5\)
\(\Leftrightarrow\frac{x^2\left(x+5\right)^2}{\left(x+5\right)^2}+\frac{25x^2}{\left(x+5\right)^2}=\frac{11\left(x+5\right)^2}{\left(x+5\right)^2}\)
Khử mẫu ta đc : \(\Leftrightarrow x^2\left(x+5\right)^2+25x^2=11\left(x+5\right)^2\)
\(\Leftrightarrow x^4+10x^3+25x^2+25x^2=11x^2+110x+275\)
\(\Leftrightarrow x^4+10x^3+50x^2-11x^2-110x-275=0\)
\(\Leftrightarrow x^4+10x^3+39x^2-110x-275=0\)
\(x-7=\left(\sqrt{x}\right)^2-\left(\sqrt{7}\right)^2=\left(\sqrt{x}-\sqrt{7}\right)\left(\sqrt{x}+\sqrt{7}\right)\)( \(x\ge0\))
\(x-6\sqrt{x}+9=\left(\sqrt{x}\right)^2-2.3.\sqrt{x}+3^2=\left(\sqrt{x}-3\right)^2\)( \(x\ge0\))
Em mới lớp 8 nên không dám chắc ạ :(
Chọn đáp án C.