For positive real numbers x,y,z so that: x+y+z = 3. Find the minimum value of expression
A = 1/( x^2 + x) + 1/(y^2+ y) +1/( z^2 +z)
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Câu 1 The function mm is defined on the real numbers by m(k) dfrac{k+2}{k+8}m(k) k+8 k+2 . What is the value of 10times m(2)10×m(2)? Answer: Câu 2 The function ff is defined on the real numbers by f(x) ax-3f(x)ax−3. What is the value of a if f(3)9f(3)9? Answer: Câu 3 The function ff is defined on the real numbers by f(x) 2x+a-3f(x)2x+a−3. What is the value of a if f(-5)11f(−5)11? Answer: Câu 4 The function ff is defined on the real numbers by f(x) 2 + x-x^2f(x)2+x−x 2 . What is the value of...
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Câu 1 The function mm is defined on the real numbers by m(k) = \dfrac{k+2}{k+8}m(k)= k+8 k+2 . What is the value of 10\times m(2)10×m(2)? Answer: Câu 2 The function ff is defined on the real numbers by f(x)= ax-3f(x)=ax−3. What is the value of a if f(3)=9f(3)=9? Answer: Câu 3 The function ff is defined on the real numbers by f(x)= 2x+a-3f(x)=2x+a−3. What is the value of a if f(-5)=11f(−5)=11? Answer: Câu 4 The function ff is defined on the real numbers by f(x) = 2 + x-x^2f(x)=2+x−x 2 . What is the value of f(-3)f(−3)? Answer: Câu 5 Given a real number aa and a function ff is defined on the real numbers by f(x)=-6\times|3x|-4f(x)=−6×∣3x∣−4. Compare: f(a)f(a) f(-a)f(−a) Câu 6 There are ordered pairs (x;y)(x;y) where xx and yy are integers such that \dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8} x 5 + 4 y = 8 1 Câu 7 Given a negative number kk and a function ff is defined on the real numbers by f(x)=\dfrac{6}{13}xf(x)= 13 6 x. Compare: f(k)f(k) f(-k)f(−k) Câu 8 Given a positive number kk and a function ff is defined on the real numbers by f(x)=\dfrac{-3}{4}x+4f(x)= 4 −3 x+4. Compare: f(k)f(k) f(-k)f(−k). Câu 9 A=(1+2+3+\ldots+90) \times(12 \times34-6 \times 68):(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6})=A=(1+2+3+…+90)×(12×34−6×68):( 3 1 + 4 1 + 5 1 + 6 1 )= Câu 10 Given that \dfrac{2x+y+z+t}{x}=\dfrac{x+2y+z+t}{y}=\dfrac{x+y+2z+t}{z}=\dfrac{x+y+z+2t}{t} x 2x+y+z+t = y x+2y+z+t = z x+y+2z+t = t x+y+z+2t . The negative value of \dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z} z+t x+y + t+x y+z + x+y z+t + y+z t+x is
20 tháng 4 2021 lúc 23:12
Let x, y, z be positive real numbers such that xy + yz + zx + xyz = 4 . Prove that :
\(3\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2\ge\left(x+2\right)\left(y+2\right)\left(z+2\right)\)
Đặt \(x=\frac{2a}{b+c};y=\frac{2b}{c+a};z=\frac{2c}{a+b}\) Thì bài toán thành chứng minh
\(3\left(\sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\right)^2\ge\frac{8\left(a+b+c\right)^3}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Áp dụng holder ta có:
\(\left(\sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\right)^2\left(2c\left(a+b\right)^2+2a\left(b+c\right)^2+2b\left(c+a\right)^2\right)\)
\(\ge\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]^3=8\left(a+b+c\right)^3\)
\(\Rightarrow VT\ge3.\frac{8\left(a+b+c\right)^3}{2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2}\)
Từ đây ta cần chứng minh:
\(3.\frac{8\left(a+b+c\right)^3}{2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2}\ge\frac{8\left(a+b+c\right)^3}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\Leftrightarrow2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2\le3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(\Leftrightarrow a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2\ge0\)( đúng )
Vậy có ĐPCM
given that (x+y):(5-z):(y+z):(9+y)=3:1:2:5
the value of x is
given that \(\dfrac{x}{x+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=x+y+z\)
Where are non- zero. The value of y is..................
my friends, help me
Sửa đề:
\(\dfrac{x}{x+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{z+y-2}\)
Dựa vào t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{x+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{z+y-2}=\dfrac{x+y+z}{x+y+x+z+z+y+\left(1+1-2\right)}=\dfrac{x+y+z}{x+x+y+y+z+z}=\dfrac{1\left(x+y+z\right)}{2\left(x+y+z\right)}=\dfrac{1}{2}\)\(x+y+z=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{y}{x+z+1}=\dfrac{1}{2}\)
\(2y=x+z+1\)
\(3y=\dfrac{1}{2}+1\)
\(y=\dfrac{1}{2}\)
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Áp dụng tính chất dãy tỉ số bằng nhau có:
\(\dfrac{x}{x+y+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=\dfrac{x+y+z}{2\left(x+y+z\right)}=\dfrac{1}{2}=x+y+z\)
\(\Rightarrow\dfrac{y}{x+z+1}=\dfrac{1}{2}\)
\(\Rightarrow2y=x+z+1\)
\(\Rightarrow3y=x+y+z+1\)
\(\Rightarrow3y=\dfrac{1}{2}+1\)
\(\Rightarrow y=\dfrac{1}{2}\)
Vậy...
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Find the maximum and minimum value of the expression
\(\frac{x+y+z}{3}+\frac{2016}{\sqrt[3]{xyz}}\)if \(x,y,z\in\left[1,2016\right]\)
Đặt \(A=\frac{x+y+z}{3}+\frac{2016}{\sqrt[3]{xyz}}\)
Tìm giá trị nhỏ nhất :Áp dụng bđt Cauchy : \(A=\frac{x+y+z}{3}+\frac{2016}{\sqrt[3]{xyz}}\ge\frac{3.\sqrt[3]{xyz}}{3}+\frac{2016}{\sqrt[3]{xyz}}\)
\(\Rightarrow A\ge\sqrt[3]{xyz}+\frac{2016}{\sqrt[3]{xyz}}\ge2\sqrt{\sqrt[3]{xyz}.\frac{2016}{\sqrt[3]{xyz}}}\)
\(\Rightarrow A\ge2\sqrt{2016}=24\sqrt{14}\) .
Dấu "=" xảy ra khi và chỉ khi \(\begin{cases}x=y=z\\\sqrt[3]{xyz}=\frac{2016}{\sqrt[3]{xyz}}\end{cases}\) \(\Leftrightarrow x=y=z=12\sqrt{14}\)
Vậy A đạt giá trị nhỏ nhất bằng \(24\sqrt{14}\) tại \(x=y=z=12\sqrt{14}\)
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Find the value of expresssion x2 + y2 + z2, if x+y+z = 5 and \(\dfrac{1}{x}\) + \(\dfrac{1}{y}\) + \(\dfrac{1}{z}\)= 0
Lời giải:
$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$
$\Rightarrow xy+yz+xz=0$
Khi đó:
$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=5^2-2.0=25$
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M.n oi dzúp mik vssssss........ :"""(
Gốc: Suppose that 2(x-3)=3(y+2) ; 5(2-z)=3(y+2) and 2x-3y+z=-4. Find the value of B=x-y+z?
Dịch: Giả sử 2(x-3)=3(y+2) ; 5(2-z)=3(y+2) và 2x-3y+z=-4. Tìm giá trị của B=x-y+z?
M.n nhớ ghi cách giải vs đáp án ra nha!
CẢM ƠN M.N RẤT RẤT NHÌU LÉM!!!! :D :3 >o<
Find the value of such that
x³+y³+z³=8?
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Find x, y, and z such that x³+y³+z³=k, for each k from 1 to 100.