Câu I:
1) So sánh hợp lí:
a)\(\left(\frac{1}{16}\right)^{200}\) và \(\left(\frac{1}{2}\right)^{1000}\)
Có: \(\left(\frac{1}{16}\right)^{200}=\left(\left(\frac{1}{2}\right)^4\right)^{200}=\left(\frac{1}{2}\right)^{800}\)
Vì \(\left(\frac{1}{2}\right)^{800}< \left(\frac{1}{2}\right)^{1000}\) nên \(\left(\frac{1}{16}\right)^{200}< \left(\frac{1}{2}\right)^{1000}\)
b) \(\left(-32\right)^{27}\) và \(\left(-18\right)^{39}\)
Có: \(\left(-32\right)^{27}=\left(\left(-32\right)^9\right)^3\);\(\left(-18\right)^{39}=\left(\left(-18\right)^{13}\right)^3\)
Có: \(\left(-32\right)^9=\left(\left(-2\right)^5\right)^9=\left(-2\right)^{45}=\left(-2\right)^{13}.2^{32}\);\(\left(-18\right)^{13}=\left(-2\right)^{13}.9^{13}=\left(-2\right)^{13}.\left(3^2\right)^{13}=\left(-2\right)^{13}.3^{26}\)
Có: \(2^{32}=\left(2^{16}\right)^2\);\(3^{26}=\left(3^{13}\right)^2\)
Có: \(2^{16}=65536;3^{13}=1594323\)
Vì \(2^{16}< 3^{13}\) nên \(2^{32}< 3^{26}\) hay \(\left(-2\right)^{13}.2^{32}>\left(-2\right)^{13}.3^{26}\)\(\Rightarrow\left(\left(-2\right)^5\right)^9>\left(-2\right)^{13}.9^{13}\)
\(\Rightarrow\left(-32\right)^9>\left(-18\right)^{13}\)\(\Rightarrow\left(-32\right)^{27}>\left(-18\right)^{39}\)
2) ĐK: x,y,z,t∈ℕ*.
*Có:\(\frac{x}{x+y+z}>\frac{x}{x+y+z+t}\);\(\frac{y}{x+y+t}>\frac{y}{x+y+z+t};\frac{z}{y+z+t}>\frac{z}{x+y+z+t}\);\(\frac{t}{x+z+t}>\frac{t}{x+y+z+t}\)
\(\Rightarrow\frac{x}{x+y+z}+\frac{y}{x+y+t}+\frac{z}{y+z+t}+\frac{t}{x+z+t}>\frac{x+y+z+t}{x+y+z+t}=1\)
Vậy M>1(1).
* Ta có: \(\frac{x}{x+y+z}< \frac{x+t}{x+y+z+t}\)
C/m: \(\Leftrightarrow\frac{x+t}{x+y+z+t}-\frac{x}{x+y+z}>0\)
\(\Leftrightarrow\frac{\left(x+t\right)\left(x+y+z\right)-x\left(x+y+z+t\right)}{\left(x+y+z+t\right)\left(x+y+z\right)}>0\)
Vì x.y,z,t>0 nên \(\left(x+y+z+t\right)\left(x+y+z\right)>0\)
\(\Rightarrow\left(x+t\right)\left(x+y+z\right)-x\left(x+y+z+t\right)>0\)
\(\Leftrightarrow x^2+xy+xz+xt+yt+zt-x^2-xy-xz-xt>0\)
\(\Leftrightarrow t\left(x+y\right)>0\)(LĐ với x,y,z,t>0)
Vậy \(\frac{x}{x+y+z}< \frac{x+t}{x+y+z+t}\).
CMTT, ta được: \(\frac{y}{x+y+t}< \frac{y+z}{x+y+z+t}\);\(\frac{z}{y+z+t}< \frac{x+z}{x+y+z+t}\);\(\frac{t}{x+z+t}< \frac{y+t}{x+y+z+t}\)
\(\Rightarrow\frac{x}{x+y+z}+\frac{y}{x+y+t}+\frac{z}{y+z+t}+\frac{t}{x+z+t}< \frac{2\left(x+y+z+t\right)}{x+y+z+t}=2\)
Vậy M<2.(2)
Từ (1) và (2), ta có: 1<M<2.
Vì 1;2 là 2 số tự nhiên liên tiếp nên M có giá trị không phải là số tự nhiên.
3) a)\(\left(3x-5\right)^{2006}+\left(y^2-1\right)^{2008}+\left(x-z\right)^{2100}=0\)
Vi \(\left(3x-5\right)^{2006};\left(y^2-1\right)^{2008};\left(x-z\right)^{2100}\ge0\) nên \(\left\{{}\begin{matrix}\left(3x-5\right)^{2006}=0\\\left(y^2-1\right)^{2008}=0\\\left(x-z\right)^{2100}=0\end{matrix}\right.\)\(\Rightarrow3x-5=y^2-1=x-z=0\)
\(\Rightarrow\left\{{}\begin{matrix}x=\frac{5}{3}\\y=\pm1\\z=\frac{5}{3}\end{matrix}\right.\)(KTM vì x,y,z∈ℕ)
Vậy không tìm được x,y,z thỏa mãn đk x,y,z ∈ℕ.
b)\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\) và \(x^2+y^2+z^2=116\)
Có:\(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}\)\(\Rightarrow\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{16}\)
Theo t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{16}=\frac{x^2+y^2+z^2}{4+9+16}=\frac{116}{29}=4\)
\(\Rightarrow x^2=4.4=16\Rightarrow\left[{}\begin{matrix}x=1\left(TM\right)\\x=-1\left(KTM\right)\end{matrix}\right.\);
\(y^2=4.9=36\Rightarrow\left[{}\begin{matrix}y=6\left(TM\right)\\y=-6\left(KTM\right)\end{matrix}\right.\);
\(z^2=4.16=64\Rightarrow\left[{}\begin{matrix}z=8\left(TM\right)\\z=-8\left(KTM\right)\end{matrix}\right.\)
Câu II:
1) a)\(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)
ĐK: \(\left\{{}\begin{matrix}2-x\ne0\\x^2-4\ne0\\2+x\ne0\\2x^2-x^3\ne0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\\x\ne0\end{matrix}\right.\)
Ta có: \(A=\frac{\left(2+x\right)^2+4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}.\frac{x^2\left(2-x\right)}{x\left(x-3\right)}\)ĐK: \(x\ne3\)
\(A=\frac{8x+4x^2}{\left(2-x\right)\left(2+x\right)}.\frac{x\left(2-x\right)}{x-3}\)
\(A=\frac{4x\left(2+x\right)x}{\left(2+x\right)\left(x-3\right)}\)\(=\frac{4x^2}{x-3}\)
b) Để A>0 thì \(\frac{4x^2}{x-3}>0\)\(\Rightarrow x-3>0\)(Vì \(4x^2>0\))
\(\Leftrightarrow x>3\)
Vậy với x>3 thì A>0.
c) Ta có: \(\left|x-7\right|=4\)
*Với \(x-7\ge0\Leftrightarrow x\ge7\)
\(\Rightarrow x-7=4\Leftrightarrow x=11\left(TM\right)\)
Thay x=11 vào A, ta có:
\(A=\frac{4x^2}{x-3}=\frac{4.11^2}{11-3}=\frac{121}{2}\)
*Với \(x-7< 0\Leftrightarrow x< 7\)
\(\Rightarrow x-7=-4\Leftrightarrow x=3\left(KTMĐKXĐ\right)\)
2) Ptích các đa thức thành nhân tử:
a)\(3x^2-7x+2=3x^2-x-6x+2\)
\(=x\left(3x-1\right)-2\left(3x-1\right)\)
\(=\left(x-2\right)\left(3x-1\right)\)
b) \(a\left(x^2+1\right)-x\left(a^2+1\right)\)
\(=ax^2+a-xa^2-x\)
\(=ax\left(x-a\right)-\left(x-a\right)\)
\(=\left(ax-1\right)\left(x-a\right)\)
Câu III:
1) Giải pt:
b)\(\sqrt{3x^2-12x+21}+\sqrt{5x^2-20x+24}=-2x^2+8x-3\)\(\left(ĐK:\left\{{}\begin{matrix}3x^2-12x+21\ge0\\5x^2-20x+24\ge0\\-2x^2+8x-3\ge0\end{matrix}\right.\right)\)
Đặt \(a=\sqrt{3x^2-12x+21};b=\sqrt{5x^2-20x+24}\left(a,b\ge0\right)\)
Ta có hpt: \(\left\{{}\begin{matrix}a^2-b^2=-2x^2+8x-3\left(1\right)\\a+b=-2x^2+8x-3\left(2\right)\end{matrix}\right.\)
Lấy (1)-(2), ta được: \(a^2-b^2-\left(a+b\right)=0\)
\(\Leftrightarrow\left(a+b\right)\left(a-b-1\right)=0\)
*Xét TH 1: a+b=0 .
\(\Rightarrow\sqrt{3x^2-12x+21}+\sqrt{5x^2-20x+24}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{3x^2-12x+21}=0\\\sqrt{5x^2-20x+24}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}3x^2-12x+21=0\\5x^2-20x+24=0\end{matrix}\right.\)(vô nghiệm)
* Xét TH 2:a-b=1
\(\Rightarrow\sqrt{3x^2-12x+21}-\sqrt{5x^2-20x+24}=1\)
\(\Leftrightarrow\sqrt{3x^2-12x+21}-3-\left(\sqrt{5x^2-20x+24}-2\right)=0\)
\(\Leftrightarrow\frac{3x^2-12x+12}{\sqrt{3x^2-12x+21}+3}-\frac{5x^2-20x+20}{\sqrt{5x^2-20x+24}+2}=0\)
\(\Leftrightarrow\left(x-2\right)^2\left[\frac{3}{\sqrt{3x^2-12x+21}+3}-\frac{5}{\sqrt{5x^2-20x+24}+2}\right]=0\)
*TH 1: \(\left(x-2\right)^2=0\Leftrightarrow x=2\left(TM\right)\)
*TH 2: \(\left[\frac{3}{\sqrt{3x^2-12x+21}+3}-\frac{5}{\sqrt{5x^2-20x+24}+2}\right]=0\)
\(\Rightarrow3\left(\sqrt{5x^2-20x+24}+2\right)-5\left(\sqrt{3x^2-12x+21}+3\right)=0\)
\(\Leftrightarrow3\sqrt{5x^2-20x+24}-5\sqrt{3x^2-12x+21}-9=0\)(vô nghiệm).
Vậy S={2}.
2) \(2^x+2^y+2^z=672\)
\(\Leftrightarrow2^x\left(1+2^{y-x}+2^{z-x}\right)=672=2^5.21\)
\(\Rightarrow x=5\left(TM\right)\);\(2^{y-5}+2^{z-5}=20\)
\(\Leftrightarrow\frac{2^y+2^z}{2^5}=20\)
\(\Rightarrow2^y\left(1+2^{z-y}\right)=640=2^7.5\)
\(\Rightarrow y=7\left(TM\right);2^{z-7}=4\)
\(\Rightarrow z-7=2\Leftrightarrow z=9\left(TM\right)\)
Vậy (x;y;z)=(5;7;9).
1) a) \(\sqrt{5-x}+\sqrt{y-2005}+\sqrt{z+2007}=\frac{1}{2}\left(x+y+z\right)\)ĐK: \(x\le4;y\ge2005;z\ge-2007\);\(x+y+z\ge0\)
Giả sử x;y;z là số nguyên.
\(\Leftrightarrow2\left(\sqrt{5-x}+\sqrt{y-2005}+\sqrt{z+2007}\right)=x+y+z\)
\(\Leftrightarrow\left(\sqrt{5-x}-1\right)^2+\left(\sqrt{y-2005}-1\right)^2+\left(\sqrt{z+2007}-1\right)^2-2\left(5-x\right)=0\)
Đặt \(a=\sqrt{5-x};b=\sqrt{y-2005};c=\sqrt{z+2007}\left(a;b;c\ge0\right)\)
\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2-2a^2=0\)
\(\Leftrightarrow-a^2-2a+1+\left(b-1\right)^2+\left(c-1\right)^2=0\)
\(\Leftrightarrow-\left(a+1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=-2\)
\(\Leftrightarrow\left(a+1\right)^2-\left(b-1\right)^2-\left(c-1\right)^2=2=2^2-1^2-1^2\)
\(\Rightarrow\left\{{}\begin{matrix}a=1\\b=2\\c=2\end{matrix}\right.\)(TM)\(\Rightarrow\left\{{}\begin{matrix}5-x=1\\y-2005=4\\z+2007=4\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=2009\\z=-2003\end{matrix}\right.\)(TM)
Vậy (x;y;z)=(4;2009;-2003).
Với x;y;z không là số nguyên thì ta không tìm được x,y,z thỏa mãn.
Vậy (x;y;z)=(4;2009;-2003).