Tìm hai số x,y biết
a/\(\dfrac{x^3}{8}=\dfrac{y^3}{27}=\dfrac{z^3}{64};x^2+2y^2-3z^2=-650\)
b/\(\dfrac{x-1}{2}=\dfrac{y+3}{4}=\dfrac{z-5}{6};5z-3x-4y=50\)
Cho x + 3y - 2z = 36. Tìm x,y,z biết
a) \(\dfrac{\text{x-1}}{\text{3}}=\dfrac{\text{y+2}}{\text{4}}=\dfrac{\text{z-2}}{\text{3}}\)
b) \(\dfrac{\text{x}}{\text{4}}=\dfrac{\text{y}}{\text{3}};\dfrac{\text{y}}{\text{2}}=\dfrac{\text{z}}{\text{5}}\)
c) 9x = 5y ; 2x = z
d) 2x = 3y = 4z
d: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+3y-2z}{\dfrac{1}{2}+3\cdot\dfrac{1}{3}-2\cdot\dfrac{1}{4}}=\dfrac{36}{1}=36\)
Do đó: x=18; y=12; z=9
Cho \(\dfrac{\text{x}}{\text{2}}=\dfrac{\text{y}}{\text{3}}=\dfrac{\text{z}}{\text{5}}\). Tìm x,y,z biết
a) x + y + z = 40
b) x - 3y + 2z = 9
c) x -y + z = 28
d) 3x + 2y = 24
a. Theo t/c của dãy tỉ số bằng nhau ta có:
x+y+z/2+3+5=40/10=4
=>x=4.2=8
=>y=4.3=12
=>z=4.5=20
b: Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-3y+2z}{2-3\cdot3+2\cdot5}=\dfrac{9}{-15}=\dfrac{-3}{5}\)
Do đó: \(\left\{{}\begin{matrix}x=-\dfrac{6}{5}\\y=\dfrac{-9}{5}\\z=-3\end{matrix}\right.\)
Tìm x,y,z biết\(\dfrac{x^2}{8}=\dfrac{y^3}{27}=\dfrac{z^3}{64};x^2+2y^2+3z^2=-650\). Nhớ giải đầy đủ nha.
Sửa đề: \(\dfrac{x^3}{8}=\dfrac{y^3}{27}=\dfrac{z^3}{64}\) và \(x^2+2y^3+3z^3=630\)
Có:\(\dfrac{x^3}{8}=\dfrac{y^3}{27}=\dfrac{z^3}{64}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
\(\Rightarrow\dfrac{x^2}{4}=\dfrac{2y^2}{18}=\dfrac{3z^2}{48}\) và \(x^2+2y^2+3z^2=630\)
Áp dụng t/c của dãy tỉ số bằng nhau có:
\(\dfrac{x^2}{4}=\dfrac{2y^2}{18}=\dfrac{3z^2}{48}=\dfrac{x^2+2y^2+3z^2}{70}=\dfrac{630}{70}=9\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=36\\y^2=\dfrac{9\cdot18}{2}=81\\z^2=\dfrac{9\cdot48}{3}=144\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\\\left[{}\begin{matrix}y=9\\y=-9\end{matrix}\right.\\\left[{}\begin{matrix}z=12\\z=-12\end{matrix}\right.\end{matrix}\right.\)
Vậy ....................
P/s: Chỗ -650 sửa thành 630 vì \(x^2+2y^2+3z^2\ge0\) nên = -650 rất vô lí --> mk sửa với lại sửa thành 630 thì kq đẹp hơn :))
~ Nếu mà đề bạn đúng thì thay số vào là đc nhé ~
Cho x,y,z >0 tm x+y+z=3
C/m :\(\dfrac{x^3}{y^3+8}+\dfrac{y^3}{z^3+8}+\dfrac{z^3}{x^3+8}\ge\dfrac{1}{9}+\dfrac{2}{27}\left(xy+yz+zx\right)\)
Cái bài này bình thường :v
Đặt \(A=\dfrac{x^3}{y^3+8}+\dfrac{y^3}{z^3+8}+\dfrac{z^3}{x^3+8}\)
\(BDT\Leftrightarrow\dfrac{x^3}{y^3+8}+\dfrac{y^3}{z^3+8}+\dfrac{z^3}{x^3+8}-\dfrac{2}{27}\left(xy+yz+xz\right)\ge\dfrac{1}{9}\)
Áp dụng BĐT AM-GM ta có:
\(\dfrac{x^3}{y^3+8}+\dfrac{y+2}{27}+\dfrac{y^2-2y+4}{27}\)
\(\ge3\sqrt[3]{\dfrac{x^3}{y^3+8}\cdot\dfrac{y+2}{27}\cdot\dfrac{y^2-2y+4}{27}}=\dfrac{x}{3}\)
Tương tự cho 2 BĐT còn lại cũng có:
\(\dfrac{y^3}{z^3+8}+\dfrac{z+2}{27}+\dfrac{z^2-2z+4}{27}\ge\dfrac{y}{3};\dfrac{z^3}{x^3+8}+\dfrac{x+2}{27}+\dfrac{x^2-2x+4}{27}\ge\dfrac{z}{3}\)
Cộng theo vế 3 BĐT trên ta có:
\(A+\dfrac{x+y+z+6}{27}+\dfrac{x^2+y^2+z^2-2\left(x+y+z\right)+12}{27}\ge\dfrac{x+y+z}{3}\)
\(\Leftrightarrow A+\dfrac{9}{27}+\dfrac{\dfrac{\left(x+y+z\right)^2}{3}+6}{27}\ge1\)\(\Leftrightarrow A\ge\dfrac{1}{3}\)
Cần chứng minh \(VT=A-\dfrac{2}{27}\left(xy+yz+xz\right)\ge\dfrac{1}{9}=VP\)
\(\Leftrightarrow VT=\dfrac{1}{3}-\dfrac{2\cdot\dfrac{\left(x+y+z\right)^2}{3}}{27}=\dfrac{1}{9}=VP\) (đúng)
Xảy ra khi \(x=y=z=1\)
P/s:Trình bày hơi khó hiểu, thông cảm :v
( Bài này làm hồi lớp 9 rồi )
Dấu " = " xảy ra khi và chỉ khi \(x=y=z=1\)
Cho 3 số thực dương x, y, z thoả mãn \(x+y+z=3\). Tìm giá trị nhỏ nhất của biểu thức: \(P=\dfrac{x^3}{y^3+8}+\dfrac{y^3}{z^3+8}+\dfrac{z^3}{x^3+8}-\dfrac{2}{27}\left(xy+yz+zx\right)\)
Tìm số x,y,z biết:
\(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\) và \(2x^2+2y^2-z^2=1\)
\(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)
\(\Leftrightarrow\left(\dfrac{x}{2}\right)^3=\left(\dfrac{y}{4}\right)^3=\left(\dfrac{z}{6}\right)^3\)
\(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)
Đặt : \(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}=k\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2k\\y=4k\\z=6k\end{matrix}\right.\)
Mà \(2x^2+2y^2-z^2=1\)
\(\Leftrightarrow2.\left(2k\right)^2+2.\left(4k\right)^2-\left(6k\right)^2=1\)
\(\Leftrightarrow8k^2+32k^2-36k^2=1\)
\(\Leftrightarrow4k^2=1\)
\(\Leftrightarrow k^2=\dfrac{1}{4}\) \(\Leftrightarrow\left[{}\begin{matrix}k=\dfrac{1}{2}\\k=-\dfrac{1}{2}\end{matrix}\right.\)
+) \(k=\dfrac{1}{2}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2.\dfrac{1}{2}=1\\y=4.\dfrac{1}{2}=2\\z=6.\dfrac{1}{2}=3\end{matrix}\right.\)
+) \(k=-\dfrac{1}{2}\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}.2=-1\\y=-\dfrac{1}{2}.4=-2\\z=-\dfrac{1}{2}.6=-3\end{matrix}\right.\)
\(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\) và \(x^2+y^2+z^2=14\)
\(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\)
=>x/2=y/4=z/6
=>x/1=y/2=z/3=k
=>x=k; y=2k; z=3k
Ta có: x^2+y^2+z^2=14
=>k^2+4k^2+9k^2=14
=>14k^2=14
=>k^2=1
TH1: k=1
=>x=1; y=2; z=3
TH2: k=-1
=>x=-1; y=-2; z=-3
Mọi người giúp em với ạ
tìm các số x,y,z biết:
a) \(\dfrac{x}{y}=\dfrac{9}{7};\dfrac{y}{z}=\dfrac{7}{3}v\text{à}x-y+z=-15\)
b) \(\dfrac{x}{y}=\dfrac{7}{20};\dfrac{y}{z}=\dfrac{5}{8}v\text{à}2x+5y-2z=100\)
c)\(5x=8y=20zv\text{à}x-y-z=3\)
d)\(\dfrac{6}{11}x=\dfrac{9}{2}y=\dfrac{18}{5}zv\text{à}-x+y+z=-120\)
a) \(\dfrac{x}{y}=\dfrac{9}{7}\)⇒\(\dfrac{x}{9}=\dfrac{y}{7}\)
\(\dfrac{y}{z}=\dfrac{7}{3}\)⇒\(\dfrac{y}{7}=\dfrac{z}{3}\)
⇒\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau,ta có:
\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=-\dfrac{15}{5}=-3\)
⇒\(\left\{{}\begin{matrix}x=-3.9=-27\\y=-3.7=-21\\z=-3.3=-9\end{matrix}\right.\)
c: Ta có: 5x=8y=20z
nên \(\dfrac{x}{\dfrac{1}{5}}=\dfrac{y}{\dfrac{1}{8}}=\dfrac{z}{\dfrac{1}{20}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{5}}=\dfrac{y}{\dfrac{1}{8}}=\dfrac{z}{\dfrac{1}{20}}=\dfrac{x-y-z}{\dfrac{1}{5}-\dfrac{1}{8}-\dfrac{1}{20}}=\dfrac{3}{\dfrac{1}{40}}=120\)
Do đó: x=24; y=15; z=6
Tìm số nguyên x, biết
a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)
b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)
\(a,-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+1}{6}=\dfrac{8}{3}\)
\(\Rightarrow-\dfrac{6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{-6x+8x+3x+3+4x+2}{12}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{9x+5}{12}=\dfrac{8}{3}\)
\(\Rightarrow27x+15=96\)
\(\Rightarrow27x=81\)
\(\Rightarrow x=3\left(tm\right)\)
\(b,\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3+5-2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow2x+1=13\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=6\left(tm\right)\)
#Toru
a) \(-\dfrac{x}{2}+\dfrac{2x}{3}+\dfrac{x+1}{4}+\dfrac{2x+2}{6}=\dfrac{8}{3}\)
\(\Rightarrow\dfrac{-6x}{12}+\dfrac{8x}{12}+\dfrac{3\left(x+1\right)}{12}+\dfrac{2\left(2x+1\right)}{12}=\dfrac{4\cdot8}{12}\)
\(\Rightarrow-6x+8x+3x+3+4x+2=32\)
\(\Rightarrow9x+5=32\)
\(\Rightarrow9x=32-5\)
\(\Rightarrow9x=27\)
\(\Rightarrow x=\dfrac{27}{9}\)
\(\Rightarrow x=3\)
b) \(\dfrac{3}{2x+1}+\dfrac{10}{4x+2}-\dfrac{6}{6x+3}=\dfrac{12}{26}\) (ĐK: \(x\ne-\dfrac{1}{2}\))
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{10}{2\left(2x+1\right)}-\dfrac{6}{3\left(2x+1\right)}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{3}{2x+1}+\dfrac{5}{2x+1}-\dfrac{2}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow\dfrac{6}{2x+1}=\dfrac{6}{13}\)
\(\Rightarrow2x+1=13\)
\(\Rightarrow2x=12\)
\(\Rightarrow x=\dfrac{12}{2}\)
\(\Rightarrow x=6\left(tm\right)\)