Câu I:
1. a) \(\left(\dfrac{1}{16}\right)^{200}=\left(\dfrac{1}{2^4}\right)^{200}=\dfrac{1}{2^{800}}\)
\(\left(\dfrac{1}{2}\right)^{1000}=\dfrac{1}{2^{1000}}\)
Vì 2800 < 21000 => \(\dfrac{1}{2^{800}}>\dfrac{1}{2^{1000}}\)
=> \(\left(\dfrac{1}{16}\right)^{200}>\dfrac{1}{2^{1000}}\)
b) (-32)27 = -(329)3 = -(245)3 = -(213.232)3 = -(213.215.217)3
(-18)39 = -(2.32)39 = -(213.326)3 = -(213.315.311)3
Vì 311 > 217 (do 177147 > 131072); 215 < 315
=> 213.215.217 < 213.315.311
=> (213.215.217)3 < (213.315.311)3
=> 3227 < 1839
=> (-32)27 < (-18)39
2. Ta có: 0 < x + y + z < x + y + z + t
=> \(\dfrac{1}{x+y+z}>\dfrac{1}{x+y+z+t}\)
=> \(\dfrac{x}{x+y+z}>\dfrac{x}{x+y+z+t}\)
Chứng minh tương tự ta có:
\(\dfrac{y}{y+z+t}>\dfrac{y}{x+y+z+t}\)
\(\dfrac{z}{z+t+x}>\dfrac{z}{x+y+z+t}\)
\(\dfrac{t}{t+x+y}>\dfrac{t}{x+y+z+t}\)
=> M > \(\dfrac{x}{x+y+z+t}+\dfrac{y}{x+y+z+t}+\dfrac{z}{x+y+z+t}+\dfrac{t}{x+y+z+y}=1\) (1)
Ta có: \(\dfrac{x}{x+y+z}< \dfrac{x+t}{x+y+z+t}\) (*)
Thật vậy, (*) <=> x(x + y + z + t) < (x + t)(x + y + z)
<=> xt + x(x + y + z) < x(x + y + z) + t(x + y + z)
<=> xt < t(x + y + z)
<=> x < x + y + z
<=> 0 < x + y + z (luôn đúng)
Tương tự ta có:
\(\dfrac{y}{y+z+t}< \dfrac{x+y}{x+y+z+t}\)
\(\dfrac{z}{z+t+x}< \dfrac{y+z}{x+y+z+t}\)
\(\dfrac{t}{t+x+y}< \dfrac{z+t}{x+y+z+t}\)
=> M < \(\dfrac{x+y}{x+y+z+t}+\dfrac{y+z}{x+y+z+t}+\dfrac{z+t}{x+y+z+t}+\dfrac{t+x}{x+y+z+t}=2\) (2)
Từ (1) và (2) => 1 < M < 2
=> M không phải số nguyên
3. a) Vì \(\left\{{}\begin{matrix}\left(3x-5\right)^{2006}\ge0\forall x\\\left(y^2-1\right)^{2008}\ge0\forall y\\\left(x-z\right)^{2100}\ge0\forall z;x\end{matrix}\right.\)
=> (3x - 5)2006 + (y2 - 1)2008 + (x - z)2100 ≥ 0
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}\left(3x-5\right)^{2006}=0\\\left(y^2-1\right)^{2008}=0\\\left(x-z\right)^{2100}=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}3x-5=0\\y^2-1=0\\x-z=0\end{matrix}\right.\)
<=> \(\left\{{}\begin{matrix}x=z=\dfrac{5}{3}\\\left[{}\begin{matrix}y=1\\y=-1\end{matrix}\right.\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}\left\{{}\begin{matrix}x=z=\dfrac{5}{3}\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=z=\dfrac{5}{3}\\y=-1\end{matrix}\right.\end{matrix}\right.\)
Vậy các cặp (x; y; z) thỏa mãn pt là (\(\dfrac{5}{3};1;\dfrac{5}{3}\)) và (\(\dfrac{5}{3};-1;\dfrac{5}{3}\))
b) Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=k\)
=> \(\left\{{}\begin{matrix}x=2k\\y=3k\\z=4k\end{matrix}\right.\)
x2 + y2 + z2 = 116
<=> (2k)2 + (3k)2 + (4k2) = 116
<=> 4k2 + 9k2 + 16k2 = 116
<=> 29k2 = 116
<=> k2 = 4
<=> \(\left[{}\begin{matrix}k=2\\k=-2\end{matrix}\right.\)
*) k = 2
=> \(\left\{{}\begin{matrix}x=2.2=4\\y=2.3=6\\z=2.4=8\end{matrix}\right.\)
*) k = -2
=> \(\left\{{}\begin{matrix}x=-2.2=-4\\y=-2.3=-6\\z=-2.4=-8\end{matrix}\right.\)
Vậy các cặp (x; y; z) thỏa mãn là (4; 6; 8) và (-4; -6; -8)
Câu II:
1. a) ĐKXĐ: \(\left\{{}\begin{matrix}2-x\ne0\\x^2-4\ne0\\2+x\ne0\\2x^2-x^3\ne0\end{matrix}\right.\) và x2 - 3x ≠ 0 <=> \(\left\{{}\begin{matrix}x\ne2\\x\ne-2\\x\ne0\\x\ne3\end{matrix}\right.\)
\(A=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{x+2}\right):\left(\dfrac{x^2-3x}{2x^2-x^3}\right)\)
\(A=\left(\dfrac{x+2}{x-2}+\dfrac{4x^2}{x^2-4}-\dfrac{x-2}{x+2}\right):\left(\dfrac{x^2-3x}{x^3-2x^2}\right)\)
\(A=\dfrac{\left(x+2\right)^2+4x^2-\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}:\dfrac{x\left(x-3\right)}{x^2\left(x-2\right)}\)
\(A=\dfrac{x^2+4x+4+4x^2-x^2+4x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x-3}{x\left(x-2\right)}\)
\(A=\dfrac{4x^2+8x}{\left(x-2\right)\left(x+2\right)}.\dfrac{x\left(x-2\right)}{x-3}\)
\(A=\dfrac{4x^2\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}\)
\(A=\dfrac{4x^2}{x-3}\)
b) \(A>0\) <=> \(\dfrac{4x^2}{x-3}>0\)
mà 4x2 > 0 ∀ x ≠ 0
=> x - 3 > 0 <=> x > 3
Kết hợp với ĐKXĐ ta có x > 3
c) |x - 7| = 4
<=> \(\left[{}\begin{matrix}x-7=4\\x-7=-4\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=11\left(TM\right)\\x=3\left(KTM\right)\end{matrix}\right.\)
<=> x = 11
Thay x = 11 (TM ĐKXĐ) vào A ta có:
\(A=\dfrac{4.11^2}{11-3}=\dfrac{484}{8}=\dfrac{121}{2}\)
2. a) 3x2 - 7x + 2 = 3x2 - 6x + x - 2 = 3x(x - 2) + (x - 2) = (3x + 1)(x -
b) a(x2 + 1) - x(a2 + 1) = ax2 + a - xa2 - x = ax(x - a) - (x - a) = (ax - 1)(x - a)
Câu III:
1. a) \(\sqrt{x-5}+\sqrt{y-2005}+\sqrt{z+2007}\le\dfrac{x-5+1+y-2005+1+z+2007+1}{2}=\dfrac{1}{2}\left(x+y+z\right)\)
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}x-5=1\\y-2005=1\\z+2007=1\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}x=6\\y=2006\\z=-2006\end{matrix}\right.\)
b) Ta có: \(\sqrt{3x^2-12x+21}+\sqrt{5x^2-20x+24}=\sqrt{3\left(x-2\right)^2+9}+\sqrt{5\left(x-2\right)^2+4}\ge\sqrt{9}+\sqrt{4}=5\)
=> VT ≥ 5
VP = -2x2 + 8x - 3 = -2(x - 2)2 + 5 ≤ 5
=> VT ≥ 5 ≥ VP
Dấu "=" xảy ra <=> \(\left\{{}\begin{matrix}VT=5\\VP=5\end{matrix}\right.\) <=> x = 2
Vậy phương trình có tập nghiệm S = {2}
2. Ta có: 29 = 512; 210 = 1024
mà 2x + 2y + 2z = 672; x < y < z
=> z ≤ 9
2x + 2y + 2z = 672 <=> 2x + 2y = 672 - 2z ≥ 672 - 29 = 160
Nếu x < y < 0 => 2x + 2y< 1 + 1 = 2 => 2x + 2y + 2z < 2 + 512 = 514 < 672
=> pt vô nghiệm
=> y > 0
=> 2y > 2x
=> 2y > \(\dfrac{2^x+2^y}{2}\) ≥ \(\dfrac{160}{2}\) = 80
=> y ≥ 7
Với 7 ≤ y < z ≤ 9 ta có các trường hợp:
TH1: y = 7; z = 8
pt <=> 2x + 27 + 28 = 672
<=> 2x = 288
=> không có x nguyên thỏa mãn
TH2: y = 7; z = 9
pt <=> 2x + 27 + 29 = 672
<=> 2x = 32
<=> x = 5 (TM)
TH3: y = 8; z = 9
pt <=> 2x + 28 + 29 = 672
<=> 2x = -96
=> không có x thỏa mãn
Vậy phương trình có cặp nghiệm nguyên (x; y; z) = (5; 7; 9)