câu hỏi 1:
\(\frac{x}{y+z+1}=\frac{y}{x+z+1}=\frac{z}{x+y-2}khidó\frac{x+y}{z+1}=?\)
Đề:
Cho các số thực x, y, z thoả mãn x + y + z = 1 và \(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1\)
\(\left(x\ne-y;y\ne-z;z\ne-x\right)\)
Giá trị của biểu thức \(P=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\) là . . .
Giải:
x + y + z = 1
=> x = 1 - (y + z)
y = 1 - (x + z)
z = 1 - (x + y)
Thay x = 1 - (y + z); y = 1 - (x + z) và z = 1 - (x + y) vào P, ta có:
\(P=\frac{x\left[1-\left(y+z\right)\right]}{y+z}+\frac{y\left[1-\left(x+z\right)\right]}{x+z}+\frac{z\left[1-\left(x+y\right)\right]}{x+y}\)
\(=\frac{x-x\left(y+z\right)}{y+z}+\frac{y-y\left(x+z\right)}{x+z}+\frac{z-z\left(x+y\right)}{x+y}\)
\(=\frac{x}{y+z}-\frac{x\left(y+z\right)}{y+z}+\frac{y}{x+z}-\frac{y\left(x+z\right)}{x+z}+\frac{z}{x+y}-\frac{z\left(x+y\right)}{x+y}\)
\(=\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)-\left(x+y+z\right)\)
\(=1-1\)
\(=0\)
ĐS: 0
Trịnh Trân Trân <3
Hay quớ ak! Mơn m nhìu nha ný! <3 <3 <3 (not thả thính =))))
a) \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
b) \(\frac{x}{y+z+1}=\frac{y}{x+z+2}=\frac{z}{x+y-2}=x+y+z\)
Tìm x ;y;z
a) \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)
\(\Rightarrow x+y+z=\frac{1}{2}\)(do 1/(x+y+z)=2)
\(\Rightarrow y+z=\frac{1}{2}-x;z+x=\frac{1}{2}-y;x+y=\frac{1}{2}-z\)
Thay vào lần lượt ta có:
\(\frac{\frac{1}{2}-x+1}{x}=2\)\(\Rightarrow x=\frac{1}{2}\)
\(\frac{\frac{1}{2}-y+2}{y}=2\)\(\Rightarrow y=\frac{5}{6}\)
\(\frac{\frac{1}{2}-z-3}{z}=2\)\(\Rightarrow z=-\frac{5}{6}\)
cho x,y,z là các số thực dương thỏa mãn\(xy+yz+zx=1\). Chứng minh rằng \(\text{x/căn(1+x^2)+y/căn(1+y^2)+z/căn(1+z^2)+1/x^2+1/y^2+1/z^2>=21/2}\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{21}{2}\)
\(\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{21}{2}\)
Đặt \(P=\frac{x}{\sqrt{1+x^2}}+\frac{y}{\sqrt{1+y^2}}+\frac{z}{\sqrt{1+z^2}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Do x,y,z là các số thực dương nên ta biến đổi \(P=\frac{1}{\sqrt{1+\frac{1}{x^2}}}+\frac{1}{\sqrt{1+\frac{1}{y^2}}}+\frac{1}{\sqrt{1+\frac{1}{z^2}}}+\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Đặt \(a=\frac{1}{x^2};b=\frac{1}{y^2};c=\frac{1}{z^2}\left(a,b,c>0\right)\)thì \(xy+yz+zx=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}=1\)và \(P=\frac{1}{\sqrt{1+a}}+\frac{1}{\sqrt{1+b}}+\frac{1}{\sqrt{1+c}}+a+b+c\)
Biến đổi biểu thức P=\(\left(\frac{1}{2\sqrt{a+1}}+\frac{1}{2\sqrt{a+1}}+\frac{a+1}{16}\right)+\left(\frac{1}{2\sqrt{b+1}}+\frac{1}{2\sqrt{b+1}}+\frac{b+1}{16}\right)\)\(+\left(\frac{1}{2\sqrt{c+1}}+\frac{1}{2\sqrt{c+1}}+\frac{c+1}{16}\right)+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{b}-\frac{3}{16}\)
Áp dụng Bất Đẳng Thức Cauchy ta có
\(P\ge3\sqrt[3]{\frac{a+1}{64\left(a+1\right)}}+3\sqrt[3]{\frac{b+1}{64\left(b+1\right)}}+3\sqrt[3]{\frac{c+1}{64\left(c+1\right)}}+\frac{15a}{16}+\frac{15b}{16}+\frac{15c}{16}-\frac{3}{16}\)
\(=\frac{33}{16}+\frac{15}{16}\left(a+b+c\right)\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{abc}\)
Mặt khác ta có \(1=\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ca}}\ge3\sqrt[3]{\frac{1}{abc}}\Leftrightarrow abc\ge27\)
\(\Rightarrow P\ge\frac{33}{16}+\frac{15}{16}\cdot3\sqrt[3]{27}=\frac{33}{16}+\frac{15}{16}\cdot9=\frac{21}{2}\)
Dấu "=" xảy ra khi a=b=c hay \(x=y=z=\frac{\sqrt{3}}{3}\)
thực hiện phép tính
a, \(\frac{x^2-yz}{1+\frac{y+x}{x}}+\frac{y^2-xz}{1+\frac{z+x}{y}}+\frac{z^2-xy}{1+\frac{x+y}{z}}\)
b, \(\left(1+\frac{y^2+z^2-x^2}{2yz}\right).\frac{1+\frac{x}{y+z}}{1-\frac{x}{y+z}}.\frac{y^2+z^2-\left(y-z\right)^2}{x+y+z}\)
c,\(\frac{2}{3}\left[\frac{1}{1+\frac{\left(2x+1\right)^2}{3}}+\frac{1}{1+\frac{\left(2x-1\right)^2}{3}}\right]\)
thực hiện phép tính
a,\(x^3+\left[\frac{x\left(2y^3-x^3\right)}{x^3+y^3}\right]^3-\left[\frac{y\left(2x^3-y^3\right)}{x^3+y^3}\right]^3\)
b,\(\frac{\frac{x\left(x+y\right)}{x-y}+\frac{x\left(x+z\right)}{x-z}}{1+\frac{\left(y-z\right)^2}{\left(x-y\right)\left(x-z\right)}}+\frac{\frac{y\left(y+z\right)}{y-z}+\frac{y\left(y+x\right)}{y-x}}{1+\frac{\left(z-x\right)^2}{\left(y-z\right)\left(y-x\right)}}+\frac{\frac{z\left(z+x\right)}{z-x}+\frac{z\left(z+y\right)}{z-y}}{1+\frac{\left(x-y\right)^2}{\left(z-x\right)\left(z-y\right)}}\)
c,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)
Giải hệ: \(\left\{{}\begin{matrix}\sqrt{x}+\sqrt{y}+\sqrt{z}-\frac{1}{\sqrt{x}}-\frac{1}{\sqrt{y}}-\frac{1}{\sqrt{z}}=\frac{8}{3}\\x+y+z+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{118}{9}\\x\sqrt{x}+y\sqrt{y}+z\sqrt{z}-\frac{1}{x\sqrt{x}}-\frac{1}{y\sqrt{y}}-\frac{1}{z\sqrt{z}}=\frac{728}{27}\end{matrix}\right.\)
ĐKXĐ: ...
Lấy pt cuối trừ 3 lần pt đầu ta được:
\(\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^3+\left(\sqrt{y}-\frac{1}{\sqrt{y}}\right)^3+\left(\sqrt{z}-\frac{1}{\sqrt{z}}\right)^3=\frac{512}{27}\)
Pt (2) tương đương:
\(x+\frac{1}{x}-2+y+\frac{1}{y}-2+z+\frac{1}{z}-2=\frac{64}{9}\)
\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2+\left(\sqrt{y}-\frac{1}{\sqrt{y}}\right)^2+\left(\sqrt{z}-\frac{1}{\sqrt{z}}\right)^2=\frac{64}{9}\)
Đặt \(\left(\sqrt{x}-\frac{1}{\sqrt{x}};\sqrt{y}-\frac{1}{\sqrt{y}};\sqrt{z}-\frac{1}{\sqrt{z}}\right)=\left(a;b;c\right)\)
Hệ trở thành:
\(\left\{{}\begin{matrix}a+b+c=\frac{8}{3}\\a^2+b^2+c^2=\frac{64}{9}\\a^3+b^3+c^3=\frac{512}{27}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b+c=\frac{8}{3}\\ab+bc+ca=0\\a^3+b^3+c^3=\frac{512}{27}\end{matrix}\right.\)
Ta có: \(a^3+b^3+c^3-3abc=\frac{512}{27}-3abc\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=\frac{512}{27}-3abc\)
\(\Leftrightarrow\frac{8}{3}.\left(\frac{64}{9}-0\right)=\frac{512}{27}-3abc\)
\(\Rightarrow abc=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b+c=\frac{8}{3}\\ab+bc+ca=0\\abc=0\end{matrix}\right.\) \(\Leftrightarrow\left(a;b;c\right)=\left(0;0;\frac{8}{3}\right)\) và hoán vị
Hay \(\left(x;y;z\right)=\left(1;1;9\right)\) và hoán vị
Cho ba số x , y , z khác 0 thỏa mãn $\frac{y+z-x}{x}$ = $\frac{z+x-y}{y}$ = $\frac{x+y-z}{z}$
Tính giá trị biểu thức P = ( 1+$\frac{x}{y}$ )( 1+$\frac{y}{z}$ )( 1+$\frac{z}{x}$ )
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y-z}{z}\\ \Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2\\ \Rightarrow\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\\ \Rightarrow x=y=z\\ \Rightarrow A=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)
\(x+y+z=1,\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1\)
tính giá trị của biểu thức A=\(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
câu 1 x,y >0 x2 + y2 =20
MinP =\(\frac{1}{x^2}\) + \(\frac{1}{y^2}\)
Câu 2 x +y + z =1
\(\frac{x}{y+z}\) +\(\frac{y}{z+x}\)+ \(\frac{z}{x+y}\)=1
Tính \(\frac{x^2}{y+z}\)+ \(\frac{y^2}{x+z}\)+ \(\frac{z^2}{x+y}\)
Nhanh vậy ta:
chơi khác kiểu không trùng ai hết.
câu 1
\(P=\frac{1}{x^2}+\frac{1}{y^2}=\frac{y^2+x^2}{\left(xy\right)^2}=\frac{20}{\left(xy\right)^2}\)(1)
Ta lại có:
\(\left(x-y\right)^2\ge0\Rightarrow x^2+y^2\ge2xy\Rightarrow xy\le\frac{20}{2}=10\)(2) Đẳng thức khi x=y
Từ (1) và (2) \(\Rightarrow P_{min}=\frac{20}{100}=\frac{1}{5}\) Khi x=y=\(\sqrt{10}\)
câu 2: Không cần đk (x+y+z)=1
\(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\) (1) =>Dk \(\hept{\begin{cases}x+z\ne0\\y+z\ne0\\x+y\ne0\end{cases}\Rightarrow\left(x+y+z\right)\ne0}\)
Nhân hai vế (1) với (x+y+z khác 0)
\(\Leftrightarrow\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\left(x+y+z\right)=1.\left(x+y+z\right)\)
\(\Leftrightarrow\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+\left(x+y+z\right)=\left(x+y+z\right)\)
\(\Rightarrow\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)=0\)
Câu 1:
Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ta có:
\(P=\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{4}{x^2+y^2}=\frac{4}{20}=\frac{1}{5}\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}x,y>0\\x^2+y^2=20\\x=y\end{cases}}\Rightarrow x=y=\sqrt{10}\)
Vậy MinP=\(\frac{1}{5}\Leftrightarrow x=y=\sqrt{10}\)
Câu 2:
Từ \(x+y+z=1\Rightarrow\hept{\begin{cases}x=1-\left(y+z\right)\\y=1-\left(x+z\right)\\z=1-\left(x+y\right)\end{cases}}\).Thay vào ta có
\(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}=\frac{x\left[1-\left(y+z\right)\right]}{y+z}+\frac{y\left[1-\left(x+z\right)\right]}{x+z}+\frac{z\left[1-\left(x+y\right)\right]}{x+y}\)
\(=\frac{x-x\left(y+z\right)}{y+z}+\frac{y-y\left(x+z\right)}{x+z}+\frac{z-z\left(x+y\right)}{x+y}\)
\(=\frac{x}{y+z}-\frac{x\left(y+z\right)}{y+z}+\frac{y}{x+z}-\frac{y\left(x+z\right)}{x+z}+\frac{z}{x+y}-\frac{z\left(x+y\right)}{x+y}\)
\(=\frac{x}{y+z}-x+\frac{y}{x+z}-y+\frac{z}{x+y}-z\)
\(=\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)-\left(x+y+z\right)=1-1=0\)
Câu 1:
Ta có:
\(\frac{1}{x^2}+\frac{x^2}{100}\ge\frac{2}{10}\left(1\right)\)
\(\frac{1}{y^2}+\frac{y^2}{100}\ge\frac{2}{10}\left(2\right)\)
Lấy (1) + (2) vế theo vế ta được
\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{x^2+y^2}{100}\ge\frac{2}{10}+\frac{2}{10}\)
\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{4}{10}-\frac{x^2+y^2}{100}=\frac{4}{10}-\frac{20}{100}=\frac{1}{5}\)
Dấu = xaey ra khi \(x^2=y^2=10\)hay \(x=y=\sqrt{10}\)
Bài 2/ Ta có:
\(\hept{\begin{cases}x+y+z=1\left(1\right)\\\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\left(2\right)\end{cases}}\)
Lấy (1) . (2) vế theo vế ta được
\(\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=1\)
\(=\frac{x^2}{y+z}+x+\frac{y^2}{z+x}+y+\frac{z^2}{x+y}+z=1\)
\(=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}+\left(x+y+z\right)=1\)
\(=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}+1=1\)
\(=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)