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Bài 1
\(x^3-2x^2+x=0\\ \Leftrightarrow x\left(x^2-2x+1\right)=0\\ \Leftrightarrow x\left(x-1\right)^2=0\)
\(\Leftrightarrow x=0\) hoặc \(\left(x-1\right)^2=0\\ \Leftrightarrow x-1=0\\ \Leftrightarrow x=1\)
\(\left(x+2\right)^2=\left(x+2\right)\left(x-2\right)\\ \Leftrightarrow\left(x+2\right)^2-\left(x+2\right)\left(x-2\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x+2-x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)4=0\\ \Leftrightarrow4x+8=0\\ \Leftrightarrow4x=-8\\ \Leftrightarrow x=-\dfrac{8}{4}\\ \Leftrightarrow x=-2\)
\(323=17.19\)
+) \(20^n+16^n-3^n-1=\left(20^n-1\right)+\left(16^n-3^n\right)\)
\(20^n-1=20^n-1^n⋮\left(20-1\right)=19\)
\(16^n-3^n⋮\left(16+3\right)=19\) (vì n chẵn)
\(\Rightarrow20^n+16^n-3^n-1⋮19\)
+) \(20^n+16^n-3^n-1=\left(20^n-3^n\right)+\left(16^n-1\right)\)
\(20^n-3^n⋮\left(20-3\right)=17\)
\(16^n-1=16^n-1^n⋮\left(16+1\right)=17\) (vì n chẵn)
\(\Rightarrow20^n+16^n-3^n-1⋮17\)
Mà \(\left(17,19\right)=1\)
\(\Rightarrow20^n+16^n-3^n-1⋮\left(17.19\right)=323\)
\(6xy-12y=6y\left(x-2\right)\\ b,5x^2-5xy-3x+3y=\left(5x^2-5xy\right)-\left(3x-3y\right)=5x\left(x-y\right)-3\left(x-y\right)=\left(x-y\right)\left(5x-3\right)\\ c,x^2-2xy-36+y^2=\left(x^2-2xy+y^2\right)-36=\left(x-y\right)^2-6^2=\left(x-y-6\right)\left(x-y+6\right)\)
\(\left(-2x+1\right)\left(2x^2+\dfrac{1}{3}x+2\right)\)
\(=-4x^3+\dfrac{2}{3}x^2-4x+2x^2+\dfrac{1}{3}x+2\)
\(=-4x^3+\dfrac{8}{3}x^2-\dfrac{11}{3}x+2\)
Câu 1: Chọn C.
Câu 2: Chọn D.
Câu 3: Chọn A.
Câu 4: Chọn A.
Câu 5: Chọn D (x=13/2).
Câu 6: Chọn A.
Câu 7: Chọn B.
Câu 8: Chọn D.
Câu 9: Chọn a.
Câu 10: Chọn d.
xét tứ giác AFCD có EA=EC;ED=EF nên tứ giác AFCD là hình bình hành
18, \(\frac{x}{2}+\frac{x^2}{8}=0\Leftrightarrow4x+x^2=0\Leftrightarrow x\left(x+4\right)=0\Leftrightarrow x=-4;x=0\)
19, \(4-x=2\left(x-4\right)^2\Leftrightarrow\left(4-x\right)-2\left(4-x\right)^2=0\)
\(\Leftrightarrow\left(4-x\right)\left[1-2\left(4-x\right)\right]=0\Leftrightarrow\left(4-x\right)\left(-7+2x\right)=0\Leftrightarrow x=4;x=\frac{7}{2}\)
20, \(\left(x^2+1\right)\left(x-2\right)+2x-4=0\Leftrightarrow\left(x^2+1\right)\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+3>0\right)=0\Leftrightarrow x=2\)
21, \(x^4-16x^2=0\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\Leftrightarrow x=0;x=\pm4\)
22, \(\left(x-5\right)^3-x+5=0\Leftrightarrow\left(x-5\right)^3-\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left[\left(x-5\right)^2-1\right]=0\Leftrightarrow\left(x-5\right)\left(x-6\right)\left(x-4\right)=0\Leftrightarrow x=4;x=5;x=6\)
23, \(5\left(x-2\right)-x^2+4=0\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5-x-2\right)=0\Leftrightarrow x=2;x=3\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x=t\)
\(\left(t+10\right)\left(t+12\right)-8=t^2+22t+120-8\)
\(=t^2+22t+112=\left(t+8\right)\left(t+14\right)\)
Theo cách đặt \(=\left(x^2+7x+8\right)\left(x^2+7x+14\right)\)
a.
\(2\left(x^2+y^2+2xy\right)+\left(x^2+2x+1\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow2\left(x+y\right)^2+\left(x+1\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+1=0\\y-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\end{matrix}\right.\)
b.
\(F=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{bc+bd}+\dfrac{c^2}{cd+ac}+\dfrac{d^2}{ad+bd}\)
\(F\ge\dfrac{\left(a+b+c+d\right)^2}{ab+ac+bc+bd+cd+ac+ad+bd}=\dfrac{\left(a+c\right)^2+\left(b+d\right)^2+2\left(a+c\right)\left(b+d\right)}{2ac+bd+\left(a+c\right)\left(b+d\right)}\)
\(F\ge\dfrac{4ac+4bd+2\left(a+c\right)\left(b+d\right)}{2ac+2bd+\left(a+c\right)\left(b+d\right)}=2\)
Dấu "=" xảy ra khi \(a=b=c=d\)