\(11\sqrt{x}+11\sqrt{y}-x-\sqrt{xy}\)
giải hpt: \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{xy}+\sqrt{y}=11+12\sqrt{13}\\x+y=134\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{xy}+\sqrt{y}=11+12\sqrt{13}\\x+y=134\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(\sqrt{x}+1\right)\left(\sqrt{y}+1\right)=12+12\sqrt{13}\\x+y=134\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x}+1=a\\\sqrt{y}+1=b\end{matrix}\right.\) \(\left(a,b>0\right)\)
\(Hpt\Leftrightarrow\left\{{}\begin{matrix}ab=12+12\sqrt{13}\\a^2+b^2-2\left(a+b\right)+2=134\end{matrix}\right.\)
\(\Leftrightarrow a^2+b^2+2ab-2\left(a+b\right)+1=134+12+12\sqrt{13}-1\)
\(\Leftrightarrow\left(a+b\right)^2-2\left(a+b\right)+1=145+12\sqrt{13}\)
\(\Leftrightarrow\left(a+b-1\right)^2=145+12\sqrt{13}\)
\(\Leftrightarrow a+b=\sqrt{145+12\sqrt{13}}+1\)
\(Hpt\Leftrightarrow\left\{{}\begin{matrix}ab=12+12\sqrt{13}\\a+b=\sqrt{145+12\sqrt{13}}+1\end{matrix}\right.\)
Số xấu quá nên dừng tại đây :D
Tìm 3 bộ số x, y, z thỏa mãn: \(\left\{{}\begin{matrix}x+y+z\le9\\\sqrt{x-1}+\sqrt{y-2}+\sqrt{z-3}+5x+4y+3z=xy+yz+xz+11\end{matrix}\right.\)
Đặt \(\left(x-1;y-2;z-3\right)=\left(a;b;c\right)=abc>0\)
Điều kiện bài toán trở thành :
\(a+1+b+2+c+3< 9\)
\(\sqrt{a+\sqrt{b}+\sqrt{c}}+\sqrt{c+5\left(a+1\right)+4\left(b+2\right)+3+\left(c+3\right)}\)
\(=\left(a+1\right)\left(b+2\right)=\left(b+2\right)\left(c+3\right)=\left(c+3\right)+\left(a+1\right)+11+a+b+c< 3\)
\(a+b+c< 3\)
\(=\sqrt{a+\sqrt{b}+\sqrt{c}+ab+bc+ca}\)
Mặt khác, do aa không âm, ta luôn có:
\(\text{(√a−1)2(a+2√a)≥0(a−1)2(a+2a)≥0}\)
\(\text{⇒a2−3a+2√a≥0⇒a2−3a+2a≥0}\)
\(\text{⇒2√a≥a(3−a)≥a(b+c)⇒2a≥a(3−a)≥a(b+c) (1)}\)
Hoàn toàn tương tự ta có:\(\text{ 2√b≥b(c+a)2b≥b(c+a) (2)}\)
\(\text{2√c≥c(a+b)2c≥c(a+b) (3)}\)
Cộng vế với vế (1);(2);(3):
\(\text{2(√a+√b+√c)≥2(ab+bc+ca)2(a+b+c)≥2(ab+bc+ca)}\)
\(\text{⇔√a+√b+√c≥ab+bc+ca⇔a+b+c≥ab+bc+ca}\)
Dấu "=" xảy ra khi và chỉ khi \(\text{a=b=c=0a=b=c=0 hoặc a=b=c=1a=b=c=1}\)
⇒x=...;y=...;z=...
Tìm x,y,z biết
\(\dfrac{\sqrt{xy}-1}{3}=\dfrac{\sqrt{yz-3}}{9}=\dfrac{\sqrt{zx-5}}{6}\) và \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=11\)
= \(\dfrac{\sqrt{xy}-1+\sqrt{yz}-3+\sqrt{zx}-5}{3+9+6}\) = \(\dfrac{11-\left(1+3+5\right)}{18}\)=\(\dfrac{1}{9}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{\sqrt{xy}-1}{3}=\dfrac{\sqrt{yz}-3}{9}=\dfrac{\sqrt{zx}-5}{6}=\dfrac{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}-1-3-5}{3+9+6}=\dfrac{11-9}{18}=\dfrac{1}{9}\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{xy}-1=\dfrac{1}{9}.3=\dfrac{1}{3}\\\sqrt{yz}-3=\dfrac{1}{9}.9=1\\\sqrt{zx}-5=\dfrac{1}{9}.6=\dfrac{2}{3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\sqrt{xy}=\dfrac{4}{3}\\\sqrt{yz}=4\\\sqrt{zx}=\dfrac{17}{3}\end{matrix}\right.\Rightarrow}\left\{{}\begin{matrix}xy=\dfrac{16}{9}\\yz=16\\zx=\dfrac{289}{9}\end{matrix}\right.\Rightarrow}\left[{}\begin{matrix}\left\{{}\begin{matrix}x=\dfrac{17}{9}\\y=\dfrac{16}{17}\\z=17\end{matrix}\right.\\\left\{{}\begin{matrix}x=-\dfrac{17}{9}\\y=-\dfrac{16}{17}\\z=-17\end{matrix}\right.\end{matrix}\right.\)
Rút gọn biểu thức
a)\(\sqrt{3}-\sqrt{2}-\sqrt{\sqrt{3}+\sqrt{2}}\)
b)\(\sqrt{11-4\sqrt{7}}-\sqrt{2}\cdot\sqrt{8+3\sqrt{7}}\)
c)\(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}\)
d)\(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{x-\sqrt{x}}\left(x>0;x\ne1\right)\)
e)\(\frac{4-4\sqrt{x}}{x-2\sqrt{x}-35}+\frac{2}{\sqrt{x}-7}-\frac{3}{\sqrt{x}+5}\left(x\ge0:x\ne49\right)\)
f)\(\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}\)
f)\(\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}:\frac{1}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}.\left(\sqrt{x}-\sqrt{y}\right)\)
\(=x-y\)
b)\(\sqrt{11-4\sqrt{7}}-\sqrt{2}.\sqrt{8+3\sqrt{7}}\)
\(=\sqrt{7-4\sqrt{7}+4}-\sqrt{16+6\sqrt{7}}\)
\(=\sqrt{\left(\sqrt{7}-2\right)^2}-\sqrt{9+6\sqrt{7}+7}\)
\(=\sqrt{7}-2-\sqrt{\left(3+\sqrt{7}\right)^2}\)(vì \(\sqrt{7}>2\))
\(=\sqrt{7}-2-3-\sqrt{7}=-5\)
c)\(\frac{x+\sqrt{xy}}{y+\sqrt{xy}}=\frac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}=\frac{\sqrt{x}}{\sqrt{y}}=\frac{\sqrt{xy}}{y}\)
d)\(\frac{\sqrt{x}}{\sqrt{x}-1}-\frac{2\sqrt{x}-1}{x-\sqrt{x}}=\frac{x-2\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{\sqrt{x}-1}{\sqrt{x}}=\frac{x-\sqrt{x}}{x}\)
giải hệ phương trình sau
\(\left\{{}\begin{matrix}x+\sqrt{y-2}+\sqrt{4-z}=y^2-5z+11\\y+\sqrt{z-2}+\sqrt{4-x}=z^2-5x+11\\z+\sqrt{x-2}+\sqrt{4-y}=x^2-5y+11\end{matrix}\right.\)
ĐKXĐ : \(2\le x,y,z\le4\)
Từ hệ phương trình ta suy ra được
\(\Sigma x+\Sigma\sqrt{x-2}+\Sigma\sqrt{4-x}=\Sigma x^2-5\Sigma x+33\\ \Leftrightarrow\Sigma\left(x^2-6x+9\right)+6=\Sigma\left(\sqrt{x-2}+\sqrt{4-x}\right)\\ \Leftrightarrow\Sigma\left(x-3\right)^2+6=\Sigma\left(\sqrt{x-2}+\sqrt{4-x}\right)\left(1\right)\)
Áp dụng bất đẳng thức \(\sqrt{A}+\sqrt{B}\le\sqrt{2\left(A+B\right)}\)
\(\Sigma\left(\sqrt{x-2}+\sqrt{4-x}\right)\le\Sigma\sqrt{2\left(x-2+4-x\right)}=\Sigma2=6\)
\(\Rightarrow\Sigma\left(x-3\right)^2+6\le6\Rightarrow\Sigma\left(x-3\right)^2\le0\)
Mà \(\Sigma\left(x-3\right)^2\ge0\)
\(\Rightarrow\left(x-3\right)^2=\left(y-3\right)^2=\left(z-3\right)^2=0\\ \Leftrightarrow x=y=z=3\)
Thay vào ta thấy thỏa mãn -> x=y=z=3 là nghiệm hpt
tìm bộ ba số (x,y,z) thỏa mãn:
\(\hept{\begin{cases}x+y+z\le9\\\sqrt{x-1}+\sqrt{y-2}+\sqrt{z-3}+5x+4y+3z=xy+yz+xz+11\end{cases}}\)
Cho các số thực x,y thỏa mãn điều kiện:
\(\sqrt{x^2+11}+\sqrt{x-2018}+x^2=\sqrt{y^2+11}+\sqrt{y-2018}+y^2\)
Tính giá trị của biểu thức: \(M=x^{11}-y^{2018}\)
trùi s ghim lên đay cx k ai giải v trùi
\(\dfrac{5}{4-\sqrt{11}}+\dfrac{1}{3+\sqrt{7}}-\dfrac{6}{\sqrt{7}-2}-\dfrac{\sqrt{7}-5}{2}=4+\sqrt{11}-3\sqrt{7}\)
\(\dfrac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}-\dfrac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}-\dfrac{y+x}{y-x}=\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(\dfrac{5\left(4+\sqrt{11}\right)}{\left(4+\sqrt{11}\right)\left(4-\sqrt{11}\right)}+\dfrac{3-\sqrt{7}}{\left(3+\sqrt{7}\right)\left(3-\sqrt{7}\right)}-\dfrac{6\left(\sqrt{7}+2\right)}{\left(\sqrt{7}-2\right)\left(\sqrt{7}+2\right)}-\dfrac{\sqrt{7}-5}{2}\)\(=\dfrac{\left(4+\sqrt{11}\right)5}{16-11}+\dfrac{3-\sqrt{7}}{9-7}-\dfrac{6\left(\sqrt{7}+2\right)}{7-4}-\dfrac{\sqrt{7}-5}{2}\)
\(=4+\sqrt{11}-\dfrac{3-\sqrt{7}}{2}-2\left(\sqrt{7}+2\right)-\dfrac{\sqrt{7}-5}{2}=\dfrac{8+2\sqrt{11}-3+\sqrt{7}-4\sqrt{7}-8-\sqrt{7}+5}{2}=\dfrac{2\sqrt{11}-4\sqrt{7}+2}{2}=1+\sqrt{11}-2\sqrt{7}\)
1) Rút gọn : \(C=\dfrac{2}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}+\dfrac{\sqrt{x}}{9-x}\)
\(D=\sqrt{xy}-\sqrt{\dfrac{x}{y}}+\sqrt{\dfrac{1}{xy}}+2\sqrt{\dfrac{y}{x}}\)
2)Cho biểu thức :
\(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
a.rút gọn
b.tìm x để A = \(\dfrac{1}{2}\)
Bài 2:
a: \(A=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+1}{\sqrt{x}+3}\)
b: Để A=1/2 thì \(\dfrac{-5\sqrt{x}+1}{\sqrt{x}+3}=\dfrac{1}{2}\)
\(\Leftrightarrow-10\sqrt{x}+2=\sqrt{x}+3\)
hay \(x\in\varnothing\)