Tìm x,y
\(\frac{-5}{\chi}\)=\(\frac{\gamma}{16}\)=\(\frac{-18}{72}\)
Tìm x,y
\(\frac{-5}{\chi}\)=\(\frac{\gamma}{16}\)=\(\frac{-18}{72}\)
\(\frac{-5}{x}=\frac{y}{16}=\frac{-18}{72}\)
tìm x và y
giúp mình với
Ta có :
-5/x = -18/72
=> -5 . 72 = -18 . x
=> -360 = -18 . x
=> x = -360 :- 18
=> x = 20
Thay x = 20 ta được :
-5/20 = y/16
=> -5 . 16 = 20 . y
=> -80 = 20 . y
=> y = -80 : 20
=> y = -4
Vậy x = 20 , y = -4
Chúc bạn học giỏi nha !!!
Dấu " / " là chia
TK mình nha !!! ^_^
x=20
y=-4
Hok tốt
^_^
Tìm x ; y biết : \(\frac{-5}{x}\)= \(\frac{y}{16}\)= \(\frac{-18}{72}\)
Xét \(\frac{-5}{x}=\frac{-18}{72}\)
\(\Rightarrow\)(-5) . 72 = x . (-18)
\(\Rightarrow\)-360 = x . (-18)
\(\Rightarrow\)x = (-360) : (-18) = 20
Xét \(\frac{y}{16}=\frac{-18}{72}\)
\(\Rightarrow\)72y = 16 . (-18)
\(\Rightarrow\)72y = -288
\(\Rightarrow\)y = (-288) : 72 = -4
Vậy x = 20 ; y = -4
Ta có :
\(\frac{-5}{x}=\frac{-18}{72}\Rightarrow\frac{-5}{x}=\frac{-1}{4}\)
\(\Rightarrow-5.4=x.-1\)
\(\Rightarrow-20=-x\)
\(\Rightarrow x=20\)
Thay x = 20 vào \(\frac{-5}{x}\)ta được :
\(\frac{-5}{20}=\frac{y}{16}\)\(\Rightarrow\frac{-1}{4}=\frac{y}{16}\Rightarrow-1.16=y.4\)
\(\Rightarrow-16=y.4\)
\(\Rightarrow y=-4\)
Vậy x = 20 , y = -4
Tk mk nha !!!
\(\frac{-5}{x}=\frac{y}{16}=\frac{-18}{72}\)
-18/72=-2/8=-4/16. Suy ra x=-4
-5/x=-2/8.Suy ra -5.8=-2.x
-40=-2x
x=20
\(\frac{-5}{x}\)=\(\frac{-y}{8}\)=\(\frac{-18}{72}\)
Yêu cầu: tìm x, y
DOGE LIKE
Ta có: \(\dfrac{-5}{x}=\dfrac{-y}{8}=\dfrac{-18}{72}\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-5}{x}=-\dfrac{18}{72}=\dfrac{-1}{4}\\\dfrac{-y}{8}=\dfrac{-18}{72}=\dfrac{-1}{4}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=20\\y=2\end{matrix}\right.\)
Vậy: (x,y)=(20;2)
\(\dfrac{-5}{x}=\dfrac{-y}{8}=\dfrac{-18}{72}\)
+)\(\dfrac{-5}{x}=\dfrac{-18}{72}\)
\(\Rightarrow x=\dfrac{-5.72}{-18}=20\)
+)\(\dfrac{-y}{8}=\dfrac{-18}{72}\)
\(\Rightarrow-y=\dfrac{-18.8}{72}=-2\)
\(\Rightarrow y=2\)
\(\frac{-5}{x}\) = \(\frac{y}{16}\)= \(\frac{-18}{72}\)
\(\frac{-5}{x}\)=\(\frac{y}{16}\)=\(\frac{-18}{72}\)
y.72 = 16.(-18)
y = -288:72
y = -4.
x.(-4) = -5.16
x = -80:(-4)
x = 20.
Vậy x =20 ; y = -4
1. Cho a,b,c > 0. Cmr: a) \(\frac{bc}{a^2+2bc}+\frac{ca}{b^2+2ca}+\frac{ab}{c^2+2ab}\le1\)
b) \(\frac{ab^2}{a^2+2b^2+c^2}+\frac{bc^2}{b^2+2c^2+a^2}+\frac{ca^2}{c^2+2a^2+b^2}\le\frac{a+b+c}{4}\)
2. Cho \(x,y,z>0;x+\frac{y}{3}+\frac{z}{5}\ge3;\frac{y}{3}+\frac{z}{5}\ge2;\frac{z}{5}\ge1.MaxP=x^2+y^2+z^2\)
3. Cho \(x>0;y\ge2;2x+y+xy\ge6.MinP=x^3+y^2\)
4. Cho \(0< \alpha< \beta< \gamma\). Giả sử x,y,z > 0 TM \(z\ge\gamma;\frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}+\frac{xyz}{\alpha\beta\gamma}=4;\frac{y}{\beta}+\frac{z}{\gamma}+\frac{yz}{\beta\gamma}=3.MinP=x^3+y^3+z^3\)
Vì đã khuya nên não cũng không còn hoạt động tốt nữa, mình làm bài 1 thôi nhé.
Bài 1:
a)
\(2\text{VT}=\sum \frac{2bc}{a^2+2bc}=\sum (1-\frac{a^2}{a^2+2bc})=3-\sum \frac{a^2}{a^2+2bc}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\sum \frac{a^2}{a^2+2bc}\geq \frac{(a+b+c)^2}{a^2+2bc+b^2+2ac+c^2+2ab}=\frac{(a+b+c)^2}{(a+b+c)^2}=1\)
Do đó: \(2\text{VT}\leq 3-1\Rightarrow \text{VT}\leq 1\) (đpcm)
Dấu "=" xảy ra khi $a=b=c$
b)
Áp dụng BĐT Cauchy-Schwarz:
\(\text{VT}=\sum \frac{ab^2}{a^2+2b^2+c^2}=\sum \frac{ab^2}{\frac{a^2+b^2+c^2}{3}+\frac{a^2+b^2+c^2}{3}+\frac{a^2+b^2+c^2}{3}+b^2}\leq \sum \frac{1}{16}\left(\frac{9ab^2}{a^2+b^2+c^2}+\frac{ab^2}{b^2}\right)\)
\(=\frac{1}{16}.\frac{9(ab^2+bc^2+ca^2)}{a^2+b^2+c^2}+\frac{a+b+c}{16}(1)\)
Áp dụng BĐT AM-GM:
\(3(ab^2+bc^2+ca^2)\leq (a^2+b^2+c^2)(a+b+c)\)
\(\Rightarrow \frac{1}{16}.\frac{9(ab^2+bc^2+ca^2)}{a^2+b^2+c^2)}\leq \frac{3}{16}(a+b+c)(2)\)
Từ $(1);(2)\Rightarrow \text{VT}\leq \frac{a+b+c}{4}$ (đpcm)
Dấu "=" xảy ra khi $a=b=c$
Bài 2/Áp dụng BĐT Bunyakovski:
\(\left(x^2+y^2+z^2\right)\left(1^2+3^2+5^2\right)\ge\left(x+3y+5z\right)^2\)
\(\Rightarrow P\ge\frac{\left(x+3y+5z\right)^2}{35}\) (*)
Ta có: \(x+3y+5z=x.1+\frac{y}{3}.9+\frac{z}{5}.25\)
\(=\frac{16z}{5}+8\left(\frac{y}{3}+\frac{z}{5}\right)+1\left(\frac{z}{5}+\frac{y}{3}+x\right)\)
\(\ge16+8.2+1.3=35\). Thay vào (*) là xong.
Đẳng thức xảy ra khi x = 1; y =3; z = 5
No choice teen, Akai Haruma, Arakawa Whiter, Phạm Lan Hương, soyeon_Tiểubàng giải, tth, Nguyễn Văn Đạt
@Nguyễn Việt Lâm
giúp em với ạ! Cần gấp lắm! Thanks nhiều!
\(\frac{-5}{x}\) = \(\frac{y}{16}\) = \(\frac{-18}{72}\)
Tìm x ; y biết :
\(\frac{-5}{x}\)= \(\frac{y}{16}\)= \(\frac{-18}{72}\)
ta có : \(\frac{y}{16}=-\frac{18}{72}\Rightarrow-18.16=y.72\Rightarrow-288=y.72\Rightarrow y=-288:72=-4\)
\(\Rightarrow-\frac{5}{x}=-\frac{4}{16}\Rightarrow x.\left(-4\right)=-5.16\Rightarrow80:\left(-4\right)=x=>x=-20\)
vậy.....