\(y=\frac{1}{x^2+\sqrt{x}}\frac{3}{5}+\frac{1}{15}+\frac{1}{39}+\frac{3}{221}+\frac{1}{119}+\frac{1}{175}\)
\(\left\{{}\begin{matrix}\frac{2}{\sqrt{x}+1}-\frac{1-x-y}{x+y}=\frac{22}{15}\\\frac{3}{\sqrt{x}+1}+\frac{5+x+y}{x+y}=3\\\\\end{matrix}\right.\)
Đặt: \(\left\{{}\begin{matrix}a=\sqrt{x}+1\\b=x+y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{a}-\frac{1-b}{b}=\frac{22}{15}\\\frac{3}{a}+\frac{5+b}{b}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{a}-\frac{1}{b}+1=\frac{22}{15}\\\frac{3}{a}+\frac{5}{b}+1=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{2}{a}-\frac{1}{b}=\frac{7}{15}\\\frac{3}{a}+\frac{5}{b}=2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{6}{a}-\frac{3}{b}=\frac{7}{5}\\\frac{6}{a}+\frac{10}{b}=4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\frac{6}{a}-\frac{3}{b}=\frac{7}{5}\\\frac{13}{b}=\frac{13}{5}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=3\\b=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3=\sqrt{x}+1\\5=x+y\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x}=2\\x+y=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5-x=1\end{matrix}\right.\)
Vậy pt có \(n_0\) \(S=\left\{4;1\right\}\)
Tính giá trị:
a, \(A=3\frac{1}{117}.\frac{1}{119}-\frac{4}{117}.5\frac{118}{119}-\frac{5}{117.119}+\frac{8}{39}\)
b, \(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...-8x^2+8x-5\) với x = 7
a, Đặt \(x=\frac{1}{117}\), \(y=\frac{1}{119}\) ta có:
\(A=\left(3+x\right)y-4x\left(5+1-y\right)-5xy+24x\)
\(=3y+xy-24x+4xy-5xy+24x\)
\(=3y\)
\(=\frac{3}{119}\)
b, Thay 8 bằng x + 1 ta có:\(B=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...-\left(x+1\right)x^2+\left(x+1\right)x-5\)
\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...-x^3-x^2+x^2+x-5\)
\(=7-5\)
= 2
a) Đặt a = \(\frac{1}{117}\)và b = \(\frac{1}{119}\)
Theo đề ta có:
A = (3 + a) b - 4a ( 5+1-b)-5ab+24a
= 3b + ab - 20a -4a + 4ab - 5ab + 24a
= 3b
= 1.\(\frac{1}{119}\) = \(\frac{3}{119}\)
Vậy A = \(\frac{3}{119}\)
TÌM GIÁ TRỊ LỚN NHẤT (có thể dùng BĐT côsi)
\(y=\left|x\right|\sqrt{25-x^2}Với-5\le x\le5\)
\(f\left(x\right)=\frac{x}{2}+\sqrt{1-x-2x^2}\)
\(E=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\)
TÍNH
\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)
\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}\)
\(\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{1+\frac{1}{3^2}+\frac{1}{4^2}}+\sqrt{1+\frac{1}{4^2}+\frac{1}{5^2}}+...+\sqrt{1+\frac{1}{2012^2}+\frac{1}{2013^2}}\)
GIÚP EM ĐI Ạ, MAI EM PHẢI KIỂM TRA RỒI
bài 1: Tính giá trị biểu thức
A = x(3x-y)-(3x+1)y tại x = 4/3; y = -1
B = \(3\frac{1}{117}.\frac{1}{119}-\frac{4}{117}.5\frac{118}{119}-\frac{8}{39}\)
Bài 2: Tìm m và n để hai đa thức đồng nhất:
f(x)=(m-1)x^2+3x+1
g(x) = x^2-nx+1
Bài 1:
Thay \(x=\frac{4}{3};y=-1\)vào biểu thức A, ta được:
\(A=\frac{4}{3}\cdot\left[3\cdot\frac{4}{3}-\left(-1\right)\right]-\left(3\cdot\frac{4}{3}+1\right)\left(-1\right)\)
\(A=\frac{20}{3}+5=\frac{35}{3}\)
Vậy khi \(x=\frac{4}{3};y=-1\)thì A=\(\frac{35}{3}\)
\(B=3\frac{1}{117}\cdot\frac{1}{119}-\frac{4}{117}\cdot5\frac{118}{119}-\frac{8}{39}\)
\(B=\frac{352}{117}\cdot\frac{1}{119}-\frac{4}{117}\cdot\frac{713}{119}-\frac{8}{39}=-\frac{412}{1071}\)
Giải các hệ PT:
a) \(\frac{1}{2x-y}+x+3y=\frac{3}{2}\) và \(\frac{4}{2x-y}-5\left(x+3y\right)=-3\)
b) \(3\left(\sqrt{x-1}\right)-\frac{4}{\sqrt{y}-1}=-1\)và \(2\left(\sqrt{x-1}\right)+\frac{3}{\sqrt{y}-1}=5\)
c) \(\frac{1}{x+y}+\sqrt{y-2}=3\)và \(\frac{-2}{x+y}+5\sqrt{y-2}=1\)
d) \(\frac{2}{\sqrt{x}-3}+\frac{1}{\sqrt{y+1}}=\frac{13}{20}\)và \(\frac{5}{\sqrt{x}-3}-\frac{2}{\sqrt{y+1}}=\frac{1}{2}\)
mấy bài này thì bạn cứ đặt ẩn phụ cho dễ nhìn hơn mà giải nhé
a, \(\hept{\begin{cases}\frac{1}{2x-y}+x+3y=\frac{3}{2}\\\frac{4}{2x-y}-5\left(x+3y\right)=-3\end{cases}}\)ĐK : \(2x\ne y\)
Đặt \(\frac{1}{2x-y}=t;x+3y=u\)hệ phương trình tương đương
\(\hept{\begin{cases}t+u=\frac{3}{2}\\4t-5u=-3\end{cases}\Leftrightarrow\hept{\begin{cases}4t+4u=6\\4t-5u=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}9u=9\\4t=-3+5u\end{cases}}\Leftrightarrow\hept{\begin{cases}u=1\\t=\frac{-3+5}{4}=\frac{1}{2}\end{cases}}}\)
Theo cách đặt \(\hept{\begin{cases}x+3y=1\\\frac{1}{2x-y}=\frac{1}{2}\end{cases}\Leftrightarrow\hept{\begin{cases}x+3y=1\\2x-y=2\end{cases}}\Leftrightarrow\hept{\begin{cases}2x+6y=2\\2x-y=2\end{cases}\Leftrightarrow}\hept{\begin{cases}7y=4\\x=\frac{y+2}{2}\end{cases}\Leftrightarrow}\hept{\begin{cases}y=\frac{4}{7}\\x=\frac{9}{7}\end{cases}}}\)
Vậy hệ pt có một nghiệm (x;y) = (9/7;4/7)
Tính
3) \(\frac{\sqrt{x}-1}{\sqrt{x}+1}+\frac{2x-\sqrt{x}-1}{x-\sqrt{x}+1}-\frac{3x\sqrt{x}-2x+\sqrt{x}-3}{x\sqrt{x}+1}\)
4) \(\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)
5)\(\frac{\sqrt{x}-1}{\sqrt{x}-3}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{x+5}{x-5\sqrt{x}+6}\)
Help !!! Mk đang cần gấp ,thank các ben
đặt \(P=\frac{1}{\sqrt{x^5-x^2+3xy+6}}+\frac{1}{\sqrt{y^5-y^2+3yz+6}}+\frac{1}{\sqrt{z^5-z^2+3zx+6}}\)
ta có:\(\left(x^3+2x^2+3x+3\right)\left(x-1\right)^2\ge0\)
\(\Leftrightarrow x^5-x^2\ge3x-3\)
cmtt=>\(y^5-y^2\ge3y-3;z^5-z^2\ge3z-3\)
\(\Rightarrow P\le\frac{1}{\sqrt{3x-3+3xy+6}}+\frac{1}{\sqrt{3y-3+3yz+6}}+\frac{1}{\sqrt{3z-3+3zx+6}}\)
\(=\frac{1}{\sqrt{3\left(x+xy+1\right)}}+\frac{1}{\sqrt{3\left(y+yz+1\right)}}+\frac{1}{\sqrt{3\left(z+zx+1\right)}}\)
áp dụng bunhia ta có:
\(3\left(x+xy+1\right)\ge\left(\sqrt{x}+\sqrt{xy}+1\right)^2\)
cmtt\(\Rightarrow P\le\frac{1}{\sqrt{x}+\sqrt{xy}+1}+\frac{1}{\sqrt{y}+\sqrt{yz}+1}+\frac{1}{\sqrt{z}+\sqrt{zx}+1}\)
đặt \(\sqrt{x}=a;\sqrt{y}=b;\sqrt{z}=c\)
\(\Rightarrow\frac{1}{\sqrt{x}+\sqrt{xy}+1}+\frac{1}{\sqrt{y}+\sqrt{yz}+1}+\frac{1}{\sqrt{z}+\sqrt{zx}+1}=\frac{1}{a+ab+1}+\frac{1}{b+bc+1}+\frac{1}{c+ca+1}\)
\(=\frac{abc}{a+ab+abc}+\frac{1}{b+bc+1}+\frac{b}{bc+abc+b}=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}=1\)
\(\Rightarrow P\le1\)
a,\(\hept{\begin{cases}\frac{1}{x-2}+\frac{1}{y-x}=1\\\frac{2}{x-2}-\frac{3}{y-1}=1\end{cases}}\) b,\(\hept{\begin{cases}\frac{7}{\sqrt{x-7}}-\frac{4}{\sqrt{y+6}}=\frac{5}{3}\\\frac{5}{\sqrt{x-y}}+\frac{3}{\sqrt{y+6}}=\frac{13}{6}\end{cases}}\)
Giải hpt này giúp em với ạ
Mình đề câu a phải như vậy nè:
\(a,\hept{\begin{cases}\frac{1}{x-2}+\frac{1}{y-1}=1\\\frac{2}{x-2}-\frac{3}{y-1}=1\end{cases}}\)\(Đkxđ:\hept{\begin{cases}x\ne2\\y\ne1\end{cases}}\)
Đặt: \(X=\frac{1}{x-2};Y=\frac{1}{y-1}\)
Ta có hệ sau:
\(\hept{\begin{cases}X+Y=1\\2X-3Y=1\end{cases}\Leftrightarrow\hept{\begin{cases}X=1-Y\\2\left(1-Y\right)-3Y=1\end{cases}}}\Leftrightarrow\hept{\begin{cases}X=1-Y\\2-5Y=1\end{cases}\Leftrightarrow\hept{\begin{cases}X=\frac{4}{5}\\Y=\frac{1}{5}\end{cases}}}\)
Với \(X=\frac{4}{5}\Rightarrow\frac{1}{x-2}=\frac{4}{5}\Leftrightarrow4\left(x-2\right)=5\Leftrightarrow x=\frac{13}{4}\)
Với \(Y=\frac{1}{5}\Rightarrow\frac{1}{y-1}=\frac{1}{5}\Leftrightarrow y-1=5\Leftrightarrow y=6\)
Vậy nghiệm của hệ pt là: \(\left(x;y\right)=\left(\frac{13}{4};6\right)\)
Câu b e nghĩ đề như vậy nè:
\(b,\hept{\begin{cases}\frac{7}{\sqrt{x-7}}-\frac{4}{\sqrt{y+6}}=\frac{5}{3}\\\frac{5}{\sqrt{x-7}}+\frac{3}{\sqrt{y+6}}=\frac{3}{6}\end{cases}}\) \(Đkxđ:\hept{\begin{cases}x>7\\x>-6\end{cases}}\)
Đặt \(\frac{1}{\sqrt{x-7}}=a\left(a>0\right);\frac{1}{\sqrt{y+6}}=b\left(b>0\right)\)
Ta có hệ pt mới: \(\hept{\begin{cases}7a-4b=\frac{5}{3}\\5a+3b=\frac{13}{6}\end{cases}}\Leftrightarrow\hept{\begin{cases}a=\frac{1}{3}\\b=\frac{1}{6}\end{cases}}\left(tmđk\right)\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{\sqrt{x-7}}=\frac{1}{3}\\\frac{1}{\sqrt{y+6}}=\frac{1}{6}\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x-7}=3\\\sqrt{y+6}=6\end{cases}}\Leftrightarrow\hept{\begin{cases}x-7=9\\x+6=36\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16\\y=30\end{cases}\left(tmđk\right)}\)
Vậy hệ pt có nghiệm \(\left(x,y\right)=\left(16;30\right)\)
Tính P = \(\frac{4+\sqrt{3}}{\sqrt{1}+\sqrt{3}}+\frac{8+\sqrt{15}}{\sqrt{3}+\sqrt{5}}+...+\frac{2n+\sqrt{n^2-1}}{\sqrt{n-1}+\sqrt{n+1}}+...+\frac{240+\sqrt{14399}}{\sqrt{119}+\sqrt{121}}\)