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Nguyễn Trần Thành ĐạtXuân Tuấn TrịnhHung nguyenHoang HungQuan Ace Legona giúp với
\(1.\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{7}+1\right)^2}-\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\dfrac{|\sqrt{7}+1|-|\sqrt{7}-1|}{\sqrt{2}}=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)
\(3a.x+1-\dfrac{x-1}{3}< x-\dfrac{2x+3}{2}+\dfrac{x}{3}+5\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)-2\left(x-1\right)}{6}< \dfrac{6x-3\left(2x+3\right)+2x+30}{6}\)
\(\Leftrightarrow6x+6-2x+2< 6x-6x-9+2x+30\)
\(\Leftrightarrow6x-2x-2x+6+2+9-30< 0\)
\(\Leftrightarrow2x-13< 0\)
\(\Leftrightarrow x< \dfrac{13}{2}\)
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 \(\left[{}\begin{matrix}x_1=4+2\sqrt{3}\\x_2=4-2\sqrt{3}\end{matrix}\right.\)
Đặt \(B=\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{n}\)
Đặt \(A=\dfrac{n-1}{1}+\dfrac{n-2}{2}+...+\dfrac{n-\left(n-2\right)}{n-2}+\dfrac{n-\left(n-1\right)}{n-1}\)
\(=\dfrac{n}{1}+\dfrac{n}{2}+...+\dfrac{n}{n-2}+\dfrac{n}{n-1}-1-1-...-1\)
\(=n+\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}-\left(n-1\right)\)
\(=\dfrac{n}{2}+\dfrac{n}{3}+...+\dfrac{n}{n-1}+\dfrac{n}{n}=n\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2}\right)=n.B\)
\(A:B=n\)
bn kt lại đề giúp mk , mk nghĩ mấu phải là x2 - 1 ; x4 - 1 ; x16 - 1
Sửa đề
\(\dfrac{1}{x-1}-\dfrac{1}{x+1}-\dfrac{2}{x^2+1}-\dfrac{4}{x^4+1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)
\(=\dfrac{2}{x^2-1}-\dfrac{2}{x^2+1}-\dfrac{4}{x^4+1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)
\(=\dfrac{4}{x^4-1}-\dfrac{4}{x^4+1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)
\(=\dfrac{8}{x^8-1}-\dfrac{8}{x^8+1}-\dfrac{16}{x^{16}+1}\)
\(=\dfrac{16}{x^{16}-1}-\dfrac{16}{x^{16}+1}=\dfrac{32}{x^{32}-1}\)
ĐKXĐ bạn tự xét nha ^^
a) \(\dfrac{3}{2y+4}-\dfrac{1}{3y+6}=\dfrac{3}{2\left(y+2\right)}-\dfrac{1}{3\left(y+2\right)}\)
\(=\dfrac{9}{6\left(y+2\right)}-\dfrac{2}{6\left(y+2\right)}=\dfrac{9-2}{6\left(y+2\right)}\)
\(=\dfrac{7}{6\left(y+2\right)}\)
b) \(\dfrac{1}{2x-3}-\dfrac{1}{2x+3}=\dfrac{2x+3}{\left(2x-3\right)\left(2x+3\right)}-\dfrac{2x-3}{\left(2x-3\right)\left(2x+3\right)}\)
\(=\dfrac{2x+3-2x+3}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{6}{4x^2-9}\)
c) \(\dfrac{1}{xy-x^2}-\dfrac{1}{y^2-xy}=\dfrac{1}{x\left(y-x\right)}-\dfrac{1}{y\left(y-x\right)}\)
\(=\dfrac{y}{xy\left(y-x\right)}-\dfrac{x}{xy\left(y-x\right)}=\dfrac{y-x}{xy\left(y-x\right)}\)
\(=\dfrac{1}{xy}\)
d) \(\dfrac{x+1}{x+4}-\dfrac{x^2-4}{x^2-16}=\dfrac{\left(x+1\right)\left(x-4\right)}{\left(x+4\right)\left(x-4\right)}-\dfrac{\left(x-2\right)\left(x+2\right)}{\left(x+4\right)\left(x-4\right)}\)
\(=\dfrac{x^2-3x-4-x^2+4}{\left(x+4\right)\left(x-4\right)}=\dfrac{-3x}{x^2-16}\)
\(S_n=\sqrt{\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{n}{4^n}}\)
\(S_{16}=\sqrt{\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{16}{4^{16}}}\)
Đặt: \(S_{16}=\sqrt{T}\Leftrightarrow T=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{16}{4^{16}}\)
\(4T=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{16}{4^{15}}\)
\(4T-T=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{16}{4^{15}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{16}{4^{16}}\right)\)
\(3T=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{15}}-\dfrac{16}{4^{16}}\)
Đặt: \(G=1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{15}}\)
\(4G=4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{14}}\)
\(4G-G=\left(4+1+\dfrac{1}{4}+...+\dfrac{1}{4^{14}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{15}}\right)\)
\(3G=4-\dfrac{1}{4^{15}}\)
\(G=\dfrac{4}{3}=\dfrac{1}{4^{15}.3}\)
\(T=\dfrac{4}{3}-\dfrac{1}{4^{15}.3}-\dfrac{16}{4^{16}}\)
\(S_{16}=\sqrt{T}=\sqrt{\dfrac{4}{3}-\dfrac{1}{4^{15}.3}-\dfrac{16}{4^{16}}}\)
bn ơi cái này mk bt lm r` sử dụng Xích - ma nha !
kq\(\simeq1,3472\)