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a)
Đặt
\(\sqrt{1+x}=a; \sqrt{1-x}=b\Rightarrow \left\{\begin{matrix} ab=\sqrt{(1+x)(1-x)}=\sqrt{1-x^2}\\ a\geq b\\ a^2+b^2=2\end{matrix}\right.\)
Khi đó:
\(A=\frac{\sqrt{1-\sqrt{1-x^2}}(\sqrt{(1+x)^3}+\sqrt{(1-x)^3})}{2-\sqrt{1-x^2}}\)
\(=\frac{\sqrt{\frac{a^2+b^2}{2}-ab}(a^3+b^3)}{a^2+b^2-ab}=\frac{\sqrt{\frac{a^2+b^2-2ab}{2}}(a+b)(a^2-ab+b^2)}{a^2+b^2-ab}\)
\(=\sqrt{\frac{a^2-2ab+b^2}{2}}(a+b)=\sqrt{\frac{(a-b)^2}{2}}(a+b)=\frac{1}{\sqrt{2}}|a-b|(a+b)\)
\(=\frac{1}{\sqrt{2}}(a-b)(a+b)=\frac{1}{\sqrt{2}}(a^2-b^2)=\frac{1}{\sqrt{2}}[(1+x)-(1-x)]=\sqrt{2}x\)
Sửa đề: \(\frac{25}{(x+z)^2}=\frac{16}{(z-y)(2x+y+z)}\)
Ta có:
Áp dụng tính chất dãy tỉ số bằng nhau thì:
\(k=\frac{a}{x+y}=\frac{5}{x+z}=\frac{a+5}{2x+y+z}=\frac{5-a}{z-y}\) ($k$ là một số biểu thị giá trị chung)
Khi đó:
\(\frac{16}{(z-y)(2x+y+z)}=\frac{25}{(x+z)^2}=(\frac{5}{x+z})^2=k^2\)
Mà: \(k^2=\frac{a+5}{2x+y+z}.\frac{5-a}{z-y}=\frac{25-a^2}{(2x+y+z)(z-y)}\)
Do đó: \(\frac{16}{(z-y)(2x+y+z)}=\frac{25-a^2}{(2x+y+z)(z-y)}\Rightarrow 16=25-a^2\)
\(\Rightarrow a^2=9\Rightarrow a=\pm 3\)
Suy ra:
\(Q=\frac{a^6-2a^5+a-2}{a^5+1}=\frac{a^5(a-2)+(a-2)}{a^5+1}=\frac{(a-2)(a^5+1)}{a^5+1}=a-2=\left[\begin{matrix}
1\\
-5\end{matrix}\right.\)
a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)
\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
\(M=\dfrac{yz\sqrt{x-1}+xz\sqrt{y-2}+xy\sqrt{z-3}}{xyz}\)
\(=\dfrac{yz\sqrt{x-1}}{xyz}+\dfrac{xz\sqrt{y-2}}{xyz}+\dfrac{xy\sqrt{z-3}}{xyz}\)
\(=\dfrac{\sqrt{x-1}}{x}+\dfrac{\sqrt{y-2}}{y}+\dfrac{\sqrt{z-3}}{z}\)
Áp dụng BĐT AM-GM ta có:
\(\sqrt{x-1}\le\dfrac{1+x-1}{2}=\dfrac{x}{2}\)\(\Rightarrow\dfrac{\sqrt{x-1}}{x}\le\dfrac{x}{2}\cdot\dfrac{1}{x}=\dfrac{1}{2}\)
\(\sqrt{y-2}=\dfrac{\sqrt{2\left(y-2\right)}}{\sqrt{2}}\le\dfrac{y}{2\sqrt{2}}\)\(\Rightarrow\dfrac{\sqrt{y-2}}{y}\le\dfrac{y}{2\sqrt{2}}\cdot\dfrac{1}{y}=\dfrac{1}{2\sqrt{2}}\)
\(\sqrt{z-3}=\dfrac{\sqrt{3\left(z-3\right)}}{\sqrt{3}}\le\dfrac{z}{2\sqrt{3}}\)\(\Rightarrow\dfrac{\sqrt{z-3}}{z}\le\dfrac{z}{2\sqrt{3}}\cdot\dfrac{1}{z}=\dfrac{1}{2\sqrt{3}}\)
Cộng theo vế 3 BĐT trên ta có:
\(M\le\dfrac{1}{2}\left(1+\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{3}}\right)\) (ĐPCM)
1/
a/ \(\sqrt{12-6\sqrt{3}}-\sqrt{21-12\sqrt{3}}\)
\(\sqrt{\left(3+\sqrt{3}\right)^2}-\sqrt{\left(3+2\sqrt{3}\right)^2}=3+\sqrt{3}-3-2\sqrt{3}=\sqrt{3}-2\sqrt{3}=-\sqrt{3}\)
b/ \(\sqrt{12}-\sqrt{27}=2\sqrt{3}-3\sqrt{3}=-\sqrt{3}\)
3/ \(C=\left(\dfrac{2x-10}{x}+\dfrac{5x+50}{x^2+5x}+\dfrac{x^2}{5x+25}\right):\dfrac{3x+15}{7}\)
\(=\left(\dfrac{2\left(x-5\right)}{x}+\dfrac{5\left(x+10\right)}{x\left(x+5\right)}+\dfrac{x^2}{5\left(x+5\right)}\right)\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\left(\dfrac{10\left(x+5\right)\left(x-5\right)}{5x\left(x+5\right)}+\dfrac{25\left(x+10\right)}{5x\left(x+5\right)}+\dfrac{x^3}{5x\left(x+5\right)}\right)\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\dfrac{10x^2-250+25x+250+x^3}{5x\left(x+5\right)}\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\dfrac{x^3+10x^2+25x}{5x\left(x+5\right)}\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\dfrac{x\left(x^2+10x+25\right)}{5x\left(x+5\right)}\cdot\dfrac{7}{3\left(x+5\right)}\)
\(=\dfrac{7\left(x+5\right)^2}{5\left(x+5\right)\cdot3\left(x+5\right)}=\dfrac{7}{15}\)
3) \(C=\left(\dfrac{2x-10}{x}+\dfrac{5x+50}{x^2+5x}+\dfrac{x^2}{5x+25}\right):\dfrac{3x+15}{7}\)
\(C=\left(\dfrac{2x-10}{x}+\dfrac{5x+50}{x\left(x+5\right)}+\dfrac{x^2}{5\left(x+5\right)}\right):\dfrac{3x+15}{7}\)
\(C=\left[\dfrac{10\left(x+5\right)\left(x-5\right)}{5x\left(x+5\right)}+\dfrac{25\left(x+10\right)}{5x\left(x+5\right)}+\dfrac{x^3}{5x\left(x+5\right)}\right]:\dfrac{3x+15}{7}\)
\(C=\left[\dfrac{10\left(x^2-25\right)+25x+250+x^3}{5x\left(x+5\right)}\right]:\dfrac{3x+15}{7}\)
\(C=\left(\dfrac{10x^2-250+25x+250-x^3}{5x\left(x+5\right)}\right).\dfrac{7}{3\left(x+5\right)}\)
\(C=\dfrac{x\left(x+2.x.5+25\right)}{5x\left(x+5\right)}.\dfrac{7}{3\left(x+5\right)}=\dfrac{x\left(x+5\right)^2}{5x\left(x+5\right)}.\dfrac{7}{3\left(x+5\right)}=\dfrac{x+5}{5}.\dfrac{7}{3\left(x+5\right)}=\dfrac{7}{15}\)
\(1.\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\dfrac{|\sqrt{7}+1|-|\sqrt{7}-1|}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
\(3a.x+1-\dfrac{x-1}{3}< x-\dfrac{2x+3}{2}+\dfrac{x}{3}+5\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)-2\left(x-1\right)}{6}< \dfrac{6x-3\left(2x+3\right)+2x+30}{6}\)
\(\Leftrightarrow6x+6-2x+2< 6x-6x-9+2x+30\)
\(\Leftrightarrow6x-2x-2x+6+2+9-30< 0\)
\(\Leftrightarrow2x-13< 0\)
\(\Leftrightarrow x< \dfrac{13}{2}\)
KL...............
\(b.5+\dfrac{x+4}{5}< x-\dfrac{x-2}{2}+\dfrac{x+3}{3}\)
\(\Leftrightarrow\dfrac{150+6\left(x+4\right)}{30}< \dfrac{30x-15\left(x-2\right)+10\left(x+3\right)}{30}\)
\(\Leftrightarrow150+6x+24< 30x-15x+30+10x+30\)
\(\Leftrightarrow6x-30x+15x-10x+150+24-30-30< 0\)
\(\Leftrightarrow-19x+114< 0\)
\(\Leftrightarrow x>6\)
KL..................
Câu 4 :
Ta có :
\(A=\dfrac{3}{1-x}+\dfrac{4}{x}\)
\(=\left(\dfrac{3}{1-x}+\dfrac{4}{x}\right)\left[\left(1-x\right)+x\right]\)
Theo BĐT Bu - nhi a - cốp xki ta có :
\(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)
\(\Leftrightarrow\left(\dfrac{3}{1-x}+\dfrac{4}{x}\right)\left[\left(1-x\right)+x\right]\ge\left(\sqrt{\dfrac{3\left(1-x\right)}{1-x}}+\sqrt{\dfrac{4x}{x}}\right)^2=\left(\sqrt{3}+2\right)^2=7+4\sqrt{3}\)
Dấu \("="\) xảy ra khi \(\dfrac{3}{\left(1-x\right)^2}=\dfrac{4}{x^2}\)
\(\Leftrightarrow3x^2=4x^2-8x+4\)
\(\Leftrightarrow x^2-8x+4=0\)
\(\Delta=64-16=48>0\)
\(\Rightarrow\left\{{}\begin{matrix}x_1=4+2\sqrt{3}\\x_2=4-2\sqrt{3}\end{matrix}\right.\)
Vậy GTNN của\(A=7+4\sqrt{3}\) khi \(\left[{}\begin{matrix}x_1=4+2\sqrt{3}\\x_2=4-2\sqrt{3}\end{matrix}\right.\)
Câu a :
Áp dụng BĐT \(\dfrac{1}{\sqrt{ab}}>\dfrac{2}{a+b}\left(a\ne b;a,b>0\right)\) ta có :
\(\dfrac{1}{\sqrt{1.1998}}>\dfrac{2}{1+1998}=\dfrac{2}{1999}\)
\(\dfrac{1}{\sqrt{2.1997}}>\dfrac{2}{2+1997}=\dfrac{2}{19999}\)
.......................................................
\(\dfrac{1}{\sqrt{1998.1}}>\dfrac{2}{1998+1}=\dfrac{2}{1999}\)
Cộng tất cả vế với nhau ta được : \(P>2.\dfrac{1998}{1999}\)
\(\Rightarrowđpcm\)
Câu a, b sao tính chất cái cuối khác những cái còn lại thế. Vậy sao biết tới đâu thì nó dừng.
\(\dfrac{\sqrt{AC^2+\left(v_n.\dfrac{t_1}{2}\right)^2}}{v}=1\)
\(\Rightarrow AC^2+v^2_n.\dfrac{t^2_1}{4}=v^2\)
\(\Rightarrow v_n^2.\dfrac{t^2_1}{4}=v^2-AC^2\)
\(\Rightarrow t_1^2=\dfrac{4\left(v^2-AC^2\right)}{v_n^2}\Rightarrow t_1=\dfrac{2\sqrt{v^2-AC^2}}{v_n}\)
Theo tao muốn chuyển đổi vậy thì :
\(\dfrac{\sqrt{AC^2+\left(v_n.\dfrac{t_1}{2}\right)^2}}{v}=\dfrac{1}{2}t_1\)
\(\Leftrightarrow\dfrac{\sqrt{AC^2+v_n^2.\dfrac{t_1^2}{4}}}{v}=\dfrac{1}{2}t_1\)
\(\Leftrightarrow\dfrac{\sqrt{\dfrac{4AC^2+\left(v_n.t_1\right)^2}{4}}}{v}=\dfrac{1}{2}t_1\)
\(\Leftrightarrow\dfrac{\sqrt{4AC^2+\left(v_n.t_1\right)^2}}{2v}=\dfrac{1}{2}t_1\)
\(\Leftrightarrow t_1=\dfrac{\sqrt{4AC^2+\left(v_n.t_1\right)^2}}{v}\)
\(\Leftrightarrow t_1v=\sqrt{4AC^2+\left(v_n.t_1\right)^2}\)
\(\Leftrightarrow t_1^2.v^2=4AC^2+v_n^2.t^2_1\)
\(\Leftrightarrow t_1^2\left(v^2-v^2_n\right)=4AC^2\)
\(\Leftrightarrow t_1^2=\dfrac{4AC^2}{v^2-v_n^2}\)
\(\Leftrightarrow t_1=\dfrac{2AC}{\sqrt{v^2-v_n^2}}\)
Are you OK??? :D