B1: a)ĐKXĐ: \(\left\{{}\begin{matrix}x-25\ne0\\x+2\sqrt{x}-15\ne0\\\sqrt{x}+5\ne0\\\sqrt{x}-3\ne0;x>0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne25\\\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)\ne0\\x>0\\\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne25\\\text{vì}\sqrt{x}+5>0\forall x\\x>0\\\sqrt{x}\ne3\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne25\\\\x>0\\x\ne9\end{matrix}\right.\)
b)\(A=\left(\dfrac{x-5\sqrt{x}}{x-25}-1\right):\left(\dfrac{25-x}{x+2\sqrt{x}-15}-\dfrac{\sqrt{x}+3}{\sqrt{x}+5}+\dfrac{\sqrt{x}-5}{\sqrt{x}-3}\right)\)
\(=\left(\dfrac{x-5\sqrt{x}-x+25}{x-25}\right):\left(\dfrac{25-x}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{x-5\sqrt{x}-x+25}{x-25}:\dfrac{25-x-x+9+x-25}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5\left(\sqrt{x}-5\right)}{x-25}:\dfrac{-x+9}{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\cdot\dfrac{\left(\sqrt{x}+5\right)\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{5}{\sqrt{x}+3}\)
c)\(A\in Z\)\(\Leftrightarrow\dfrac{5}{\sqrt{x}+3}\in Z\)
Hay \(5⋮\sqrt{x}+3\) hay \(\sqrt{x}+3\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)
\(\Leftrightarrow\sqrt{x}=2\left(x>0\right)\)\(\Leftrightarrow x=4\) (thỏa mãn đkxđ)
Vậy x=4 thì A nhận giá trị nguyên
d)\(B=\dfrac{\dfrac{5}{\sqrt{x}+3}\left(x+16\right)}{5}=\dfrac{x+16}{\sqrt{x}+3}\)
Đặt \(\sqrt{x}=a>0\) thì ta có:
\(B=\dfrac{a^2+16}{a+3}\)\(\Leftrightarrow-a^2+Ba+3B-16=0\)
\(\Delta=B^2-4\left(-1\right)\left(3B-16\right)=\left(B-4\right)\left(B+16\right)\)
pt có nghiệm khi \(\Delta\ge0\Leftrightarrow\left(B-4\right)\left(B+16\right)\ge0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}B-4\ge0\\B+16\ge0\end{matrix}\right.\\\left\{{}\begin{matrix}B-4\le0\\B+16\le0\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}B\ge4\\B\le-16\end{matrix}\right.\)
Dấu "=" xảy ra khi \(a=2\)(thỏa mãn)\(\Leftrightarrow x=4\)(thỏa mãn)
Vậy MinA=4 khi x=4
B2: a)\(\left(x^2-x+1\right)\left(x^2+4x+1\right)=6x^2\)
\(\Leftrightarrow x^4+3x^3-2x^2+3x+1=6x^2\)
\(\Leftrightarrow x^4+3x^3-8x^2+3x+1=0\)
\(\Leftrightarrow x^2\left(x^2+5x+1\right)-2x\left(x^2+5x+1\right)+\left(x^2+5x+1\right)=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)\left(x^2+5x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2+5x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+5x+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\\left(x+\dfrac{5}{2}\right)^2=\dfrac{21}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+\dfrac{5}{2}=\pm\dfrac{\sqrt{21}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\pm\dfrac{\sqrt{21}}{2}-\dfrac{5}{2}\end{matrix}\right.\)
Vậy tập nghiệm của pt là \(S=\left\{\pm\dfrac{\sqrt{21}}{2}-\dfrac{5}{2};1\right\}\)
b)\(3x^2+2x=2\sqrt{x^2+x}+1-x\)
ĐK:\(x\ge0\)
\(\Leftrightarrow3x^2+3x-1=2\sqrt{x^2+x}\)
\(\Leftrightarrow9x^4+18x^3+3x^2-6x+1=4\left(x^2+x\right)\)
\(\Leftrightarrow9x^4+18x^3-x^2-10x+1=0\)
\(\Leftrightarrow x^2\left(9x^2+9x-1\right)+x\left(9x^2+9x-1\right)-\left(9x^2+9x-1\right)=0\)
\(\Leftrightarrow\left(x^2+x-1\right)\left(9x^2+9x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x-1=0\\9x^2+9x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\dfrac{1}{2}\right)^2=\dfrac{5}{4}\\9\left(x+\dfrac{1}{2}\right)^2=\dfrac{13}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\pm\dfrac{\sqrt{5}}{2}\\x+\dfrac{1}{2}=\pm\dfrac{\sqrt{13}}{6}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\dfrac{\sqrt{5}}{2}-\dfrac{1}{2}\\x=\pm\dfrac{\sqrt{13}}{6}-\dfrac{1}{2}\end{matrix}\right.\)
Thay vào pt ta có \(x=\pm\dfrac{\sqrt{5}}{2}-\dfrac{1}{2}\) (thỏa mãn)
Vậy tập nghiệm của pt là \(S=\text{}\left\{\pm\dfrac{\sqrt{5}}{2}-\dfrac{1}{2}\right\}\)
c)\(\sqrt{x+3}+2x\sqrt{x+1}=2x+\sqrt{x^2+4x+3}\)
ĐK:\(x\ge -1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+3}=a\\\sqrt{x+1}=b\end{matrix}\right.\)\((a;b\ge 0)\) thì ta có:
\(pt\Leftrightarrow a+2xb=2x+a\text{}b\)
\(\Leftrightarrow b\left(2x-a\right)-\left(2x-a\right)=0\)
\(\Leftrightarrow\left(b-1\right)\left(2x-a\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=1\\2x=\sqrt{x+3}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x+1=1\\4x^2=x+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left(4x+3\right)\left(x-1\right)=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\) (thỏa mãn,thay \(x=-\dfrac{3}{4}\) vào pt và vô nghiệm)
Vậy tập nghiệm của pt là \(S=\left\{0;1\right\}\)
d)\(x^2+9x+20=2\sqrt{3x+10}\)
ĐK:\(x\ge-\dfrac{10}{3}\)
\(pt\Leftrightarrow x^4+18x^3+121x^2+360x+400=4\left(3x+10\right)\)
\(\Leftrightarrow x^4+18x^3+121x^2+348x+360=0\)
\(\Leftrightarrow x^2\left(x^2+12x+40\right)+6x\left(x^2+12x+40\right)+9\left(x^2+12x+40\right)=0\)
\(\Leftrightarrow\left(x^2+6x+9\right)\left(x^2+12x+40\right)=0\)
\(\Leftrightarrow\left(x+3\right)^2\left(x^2+12x+40\right)=0\)
Dễ thấy: \(x^2+12x+40=\left(x+6\right)^2+4>0\forall x\)
Hay x+3=0<=>x=-3
Vậy tập nghiệm của pt là S={-3}
B3:a) \(VT=\dfrac{\sqrt{\dfrac{abc+4}{a}-4\sqrt{\dfrac{bc}{a}}}}{\sqrt{abc}-2}=\dfrac{\sqrt{\dfrac{abc+4}{a}-\dfrac{4\sqrt{abc}}{a}}}{\sqrt{abc}-2}\)
\(=\dfrac{\sqrt{\dfrac{abc+4-4\sqrt{abc}}{a}}}{\sqrt{abc}-2}=\dfrac{\sqrt{\dfrac{\left(\sqrt{abc}-2\right)^2}{a}}}{\sqrt{abc}-2}\)
\(=\dfrac{\dfrac{\sqrt{abc}-2}{\sqrt{a}}}{\sqrt{abc}-2}\left(\text{vì }\sqrt{abc}>2\Leftrightarrow\sqrt{abc}-2>0\right)\)\(=\dfrac{1}{\sqrt{a}}=VP\)
b)Đặt \(\left\{{}\begin{matrix}a+b+c=3u\\ab+bc+ca=3v^2\\abc=w^3\end{matrix}\right.\). Khi đó ta có:
\(f(w^3)=-8w^6+A(u,v^2)w^3+B(u,v^2)\ge 0\)
\(f\) là một hàm lõm, nên ta có thể nói BĐT cho một giá trị cực trị của \(w^3\) xảy ra hai trường hợp
*)\(w^3=0\) vì vai trò của a;b;c như nhau nên ta giả sử \(c=0\)và \(gt\Rightarrow ab=1\) khi đó ta cần chứng minh
\(\dfrac{1}{2a+2\cdot0\cdot c+1}+\dfrac{1}{2b+2\cdot0\cdot a+1}+\dfrac{1}{2\cdot0+2\cdot1+1}\ge1\)
\(\Leftrightarrow\dfrac{1}{2a+1}+\dfrac{1}{2b+1}+\dfrac{1}{3}\ge1\)
\(\Leftrightarrow\dfrac{a+b+1}{2\left(a+b\right)+5}\ge\dfrac{1}{3}\)\(\Leftrightarrow a+b\ge2\)
BĐT cuối đúng theo BĐT AM-GM \(a+b\ge2\sqrt{ab}=2\)
*)2 biến bằng nhau ko mất tính tổng quát giả sử \(a=b\) và \(gt\Rightarrow c=\dfrac{1-a^2}{2a}\left(0< a< 1;a\ne0\right)\)
Khi đó ta cần chứng minh \(\dfrac{1}{2a+2a\cdot\dfrac{1-a^2}{2a}+1}+\dfrac{1}{2a+2\cdot\dfrac{1-a^2}{2a}a+1}+\dfrac{1}{2\dfrac{1-a^2}{2a}+2a^2+1}\ge1\)
\(\Leftrightarrow\dfrac{2}{-a^2+2a+2}+\dfrac{a}{2a^3-a^2+a+1}\ge1\)
\(\Leftrightarrow-\dfrac{3a^3+4a+2}{\left(2a+1\right)\left(a^2-2a-2\right)\left(a^2-a+1\right)}-1\ge0\)
\(\Leftrightarrow\dfrac{a^2(1-a)(1+3a-2a^2)}{\left(2a+1\right)\left(a^2-2a-2\right)\left(a^2-a+1\right)}\ge0\)
\(\Leftrightarrow a^2(1-a)(1+3a-2a^2)\ge0\)*luôn đúng vì \(\left(0< a< 1;a\ne0\right)\) *
Dấu "=" xảy ra khi \(\left(a;b;c\right)=\left(0;1;1\right)\) và hoán vị
B4:
a)Xét t.giác vuông AHB (vuông tại H (AH dg cao(gt))
HD vuông góc AB (gt)
AH2=AD*AB (hệ thức lượng) (1)
Xét t.giác vuông AHC(vuông tại H (AH dg cao(gt))
HE vuông góc AC (gt)
AH2=AE*AC (hệ thức lượng) (2)
Từ (1) và (2) ta có \(\Leftrightarrow\dfrac{AD}{AE}=\dfrac{AC}{AB}\)
Xét t.giác AED và t.giác ABC: DAE=CAB=90 độ (t.giác ABC vuông tại A(gt))
\(\dfrac{AD}{AE}=\dfrac{AC}{AB}\)
suy ra t.giác AED ~ t.giác ABC (c.g.c)
b)AD*AB=AE*AC (cmt) suy ra \(\dfrac{AD}{AC}=\dfrac{AE}{AB}\)
\(\Leftrightarrow\left(\dfrac{AD}{AC}\right)^2=\dfrac{AD\cdot AE}{AC\cdot AB}\)
Lai co:\(SADHE=\dfrac{1}{2}SABC\Rightarrow AD\cdot AE=\dfrac{AB\cdot AC}{4}\)
\(\Rightarrow\dfrac{AD\cdot AE}{AB\cdot AC}=\dfrac{1}{4}\)\(\Rightarrow\left(\dfrac{AD}{AC}\right)^2=\dfrac{1}{4}\Leftrightarrow\dfrac{AD}{AC}=\dfrac{1}{2}\)
Hay tam giác ABC là tam giác vuông cân
B5:Ta có:
\(x\sqrt{\dfrac{\left(2011+y^2\right)\left(2011+z^2\right)}{2011+x^2}}=x\sqrt{\dfrac{\left(xy+yz+xz+y^2\right)\left(xy+yz+xz+z^2\right)}{xy+yz+xz+x^2}}\)
\(=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)\(=x\sqrt{\left(y+z\right)^2}\)\(=x\left(y+z\right)\) (vì x;y;z là các số dương)
Tương tự ta có: \(y\sqrt{\dfrac{\left(2011+x^2\right)\left(2011+z^2\right)}{\left(2011+y^2\right)}}=y\left(x+z\right)\) và \(z\sqrt{\dfrac{\left(2011+y^2\right)\left(2011+x^2\right)}{\left(2011+z^2\right)}}=z\left(x+y\right)\)
Cộng theo vế 3 đẳng thức trên ta có: \(Q=2\left(xy+yz+xz\right)=2\cdot2011=4022\)
Vậy Q=4022