\(\dfrac{x}{2}\)=\(\dfrac{y}{3}\);\(\dfrac{y}{5}\)=\(\dfrac{z}{7}\)và x+y+z=92
Những câu hỏi liên quan
\(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}\)và 2x+y-z=81
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}\)và 5x-y+3z=124
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)và x.y.z=810
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}\)và\(x^2.y^2.z^2=288^2\)
a.
Đặt \(\dfrac{x}{5}=\dfrac{y}{3}=\dfrac{z}{4}=k\Rightarrow\left\{{}\begin{matrix}x=5k\\y=3k\\z=4k\end{matrix}\right.\)
Thế vào \(2x+y-z=81\)
\(\Rightarrow2.5k+3k-4k=81\)
\(\Rightarrow9k=81\)
\(\Rightarrow k=9\)
\(\Rightarrow\left\{{}\begin{matrix}x=5k=45\\y=3k=27\\z=4k=36\end{matrix}\right.\)
b.
Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{2}=k\Rightarrow\left\{{}\begin{matrix}x=3k\\y=5k\\z=2k\end{matrix}\right.\)
Thế vào \(5x-y+3z=124\)
\(\Rightarrow5.3k-5k+3.2k=124\)
\(\Rightarrow16k=124\)
\(\Rightarrow k=\dfrac{31}{4}\) \(\Rightarrow\left\{{}\begin{matrix}x=3k=\dfrac{93}{4}\\y=5k=\dfrac{155}{4}\\z=2k=\dfrac{31}{2}\end{matrix}\right.\)
Đúng 0
Bình luận (0)
c.
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=5k\end{matrix}\right.\)
Thế vào \(xyz=810\)
\(\Rightarrow2k.3k.5k=810\)
\(\Rightarrow k^3=27\)
\(\Rightarrow k=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k=6\\y=3k=9\\z=5k=15\end{matrix}\right.\)
Đúng 0
Bình luận (0)
d.
Đặt \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{6}=k\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k\\y=3k\\z=6k\end{matrix}\right.\)
Thế vào \(x^2y^2z^2=288^2\)
\(\Rightarrow\left(2k\right)^2.\left(3k\right)^2.\left(6k\right)^2=288^2\)
\(\Rightarrow\left(k^2\right)^3=64\)
\(\Rightarrow k^2=4\)
\(\Rightarrow k=\pm2\)
\(\Rightarrow\left\{{}\begin{matrix}x=2k=4\\y=3k=6\\z=6k=12\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x=2k=-4\\y=3k=-6\\z=6k=-12\end{matrix}\right.\)
Đúng 0
Bình luận (0)
Tìm x, y, z
dfrac{x+y+2}{z}dfrac{y+z+1}{x}dfrac{z+x-3}{y}dfrac{1}{x+y+z}
Áp dụng tích chất của dãy tỉ số bằng nhau, ta có
dfrac{x+y+2}{z}dfrac{y+z+1}{x}dfrac{z+x-3}{y}
dfrac{x+y+2+y+z+1+z+x-3}{z+x+y}dfrac{2left(x+y+zright)+left(1+2-3right)}{z+x+y}2
Vìdfrac{x+y+2}{z}dfrac{y+z+1}{x}dfrac{z+x-3}{y}dfrac{1}{x+y+z}
2dfrac{1}{x+y+z}2left(x+y+zright)1x+y+zdfrac{1}{2}
dfrac{x+y+2}{z}2x+y+22z
dfrac{y+z+1}{x}2y+z+12x
dfrac{z+x-3}{y}2z+x-32y
dfrac{1}{x+y+z}2x+y+zdfrac{1}{2}
+) x+y+z dfrac{1}{2}y+zdfra...
Đọc tiếp
Tìm x, y, z
\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\)
Áp dụng tích chất của dãy tỉ số bằng nhau, ta có
\(\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}\\ =\dfrac{x+y+2+y+z+1+z+x-3}{z+x+y}=\dfrac{2\left(x+y+z\right)+\left(1+2-3\right)}{z+x+y}=2\\ Vì\dfrac{x+y+2}{z}=\dfrac{y+z+1}{x}=\dfrac{z+x-3}{y}=\dfrac{1}{x+y+z}\\ =>2=\dfrac{1}{x+y+z}=>2\left(x+y+z\right)=1=>x+y+z=\dfrac{1}{2}\\ =>\dfrac{x+y+2}{z}=2=>x+y+2=2z\\ \dfrac{y+z+1}{x}=2=>y+z+1=2x\\ \dfrac{z+x-3}{y}=2=>z+x-3=2y\\ \dfrac{1}{x+y+z}=2=>x+y+z=\dfrac{1}{2}\)
+) x+y+z = \(\dfrac{1}{2}=>y+z=\dfrac{1}{2}-x=>\dfrac{1}{2}-x+1=2x=>3x=\dfrac{3}{2}=>x=\dfrac{1}{2}\)
+)\(x+y+z=\dfrac{1}{2}=>x+y=\dfrac{1}{2}-z=>\dfrac{1}{2}-z+2=2z=>3z=\dfrac{5}{2}=>z=\dfrac{5}{6}\)
\(=>x+y+z=\dfrac{1}{2}+\dfrac{5}{6}+y=\dfrac{1}{2}=>\dfrac{4}{3}+y=\dfrac{1}{2}=>y=\dfrac{-5}{6}\)
Vậy \(x=\dfrac{1}{2}\\ y=\dfrac{-5}{6}\\ z=\dfrac{5}{6}\)
Ê mấy bọn 7B Nguyễn Lương Bằng ơi bài 2 Toán chiều làm thế này đúng chưa! Góp ý nha!
Xem thêm câu trả lời
\(\dfrac{x^3}{y^3}+\dfrac{y^3}{z^3}+\dfrac{z^3}{x^3}\ge\dfrac{x^2}{y^2}+\dfrac{y^2}{z^2}+\dfrac{z^2}{x^2}\)
Cần thêm điều kiện x;y;z dương, nếu không đây là 1 BĐT sai
Đúng 0
Bình luận (0)
giải các hệ phương trìnha dfrac{5}{x-1}+dfrac{1}{y-1}10dfrac{1}{x-1}-dfrac{3}{y-1}18b dfrac{5}{x+y-3}-dfrac{2}{x-y+1}8dfrac{3}{x+y-3}+dfrac{1}{x-y+1}dfrac{3}{2}c sqrt{x-1}-3sqrt{y+2}22sqrt{x-1}+5sqrt{y+2}15d dfrac{7}{sqrt{x-7}}-dfrac{4}{sqrt{y+6}}dfrac{5}{3}dfrac{5}{sqrt{x-7}}+dfrac{3}{sqrt{y+6}}dfrac{13}{6}e 7x^2+13y-395x^2-11y33f 2left(x-1right)^2-3y^375left(x-1right)^2+6y^34
Đọc tiếp
giải các hệ phương trình
a \(\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\)
\(\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\)
b \(\dfrac{5}{x+y-3}-\dfrac{2}{x-y+1}=8\)
\(\dfrac{3}{x+y-3}+\dfrac{1}{x-y+1}=\dfrac{3}{2}\)
c \(\sqrt{x-1}-3\sqrt{y+2}=2\)
\(2\sqrt{x-1}+5\sqrt{y+2}=15\)
d \(\dfrac{7}{\sqrt{x-7}}-\dfrac{4}{\sqrt{y+6}}=\dfrac{5}{3}\)
\(\dfrac{5}{\sqrt{x-7}}+\dfrac{3}{\sqrt{y+6}}=\dfrac{13}{6}\)
e \(7x^2+13y=-39\)
\(5x^2-11y=33\)
f \(2\left(x-1\right)^2-3y^3=7\)
\(5\left(x-1\right)^2+6y^3=4\)
a) Ta có: \(\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-1}+\dfrac{1}{y-1}=10\\\dfrac{5}{x-1}-\dfrac{15}{y-1}=90\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{16}{y-1}=-80\\\dfrac{1}{x-1}-\dfrac{3}{y-1}=18\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{-1}{5}\\\dfrac{1}{x-1}=18+\dfrac{3}{y-1}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{4}{5}\\x-1=\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{4}{3}\\y=\dfrac{4}{5}\end{matrix}\right.\)
Đúng 1
Bình luận (0)
Rút gọn:
a) A= \(\dfrac{x+y}{x-y}-\dfrac{x}{x+y}+\dfrac{2y^2}{x^2-y^2}\)
b) B= \(\dfrac{x}{x-2}-\dfrac{10}{\left(x-2\right)\left(x+3\right)}-\dfrac{x-1}{x+3}\)
c) C= \(\dfrac{1}{x-1}-\dfrac{x-1}{x^2+x+1}-\dfrac{3}{x^3-1}\)
a: \(A=\dfrac{x^2+2xy+y^2-x^2+xy+2y^2}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{3y^2+3xy}{\left(x-y\right)\left(x+y\right)}=\dfrac{3y}{x-y}\)
Đúng 0
Bình luận (0)
Tìm x,y,z biết:
a) \(\dfrac{x}{5}=\dfrac{y}{2}\) và \(x-y=9\)
b) \(\dfrac{x-3}{12}=\dfrac{-3}{3-x}\)
c) \(\dfrac{x}{2}=\dfrac{y}{3};\dfrac{y}{5}=\dfrac{z}{4}\) và \(x-y-z=-49\)
a: Áp dụng tính chất của DTSBN, ta được:
x/5=y/2=(x-y)/(5-2)=9/3=3
=>x=15; y=6
b: =>(x-3)/12=3/(x-3)
=>(x-3)^2=36
=>(x-9)(x+3)=0
=>x=9 hoặc x=-3
c; x/2=y/3
=>x/10=y/15
y/5=z/4
=>y/15=z/12
=>x/10=y/15=z/12=(x-y-z)/(10-15-12)=-49/-17=49/17
=>x=490/17; y=735/17; z=588/17
Đúng 2
Bình luận (0)
Rút gọn các biểu thức:a) {dfrac{1}{x^2} + dfrac{1}{y^2} + dfrac{2}{x+y}(dfrac{1}{x} + dfrac{1}{y})} : dfrac{x^3+y^3}{x^2y^2}b) {dfrac{1}{left(2x-yright)^2} + dfrac{2}{4x^2-y^2} + dfrac{1}{left(2x+yright)^2}} . dfrac{4x^2+4xy+y^2}{16x}c) (dfrac{x^2-xy}{x^2y+y^3} - dfrac{2x^2}{y^3-xy^2+x^2y-x^3})(1 - dfrac{y-1}{x} - dfrac{y}{x^2})
Đọc tiếp
Rút gọn các biểu thức:
a) {\(\dfrac{1}{x^2}\) + \(\dfrac{1}{y^2}\) + \(\dfrac{2}{x+y}\)(\(\dfrac{1}{x}\) + \(\dfrac{1}{y}\))} : \(\dfrac{x^3+y^3}{x^2y^2}\)
b) {\(\dfrac{1}{\left(2x-y\right)^2}\) + \(\dfrac{2}{4x^2-y^2}\) + \(\dfrac{1}{\left(2x+y\right)^2}\)} . \(\dfrac{4x^2+4xy+y^2}{16x}\)
c) (\(\dfrac{x^2-xy}{x^2y+y^3}\) - \(\dfrac{2x^2}{y^3-xy^2+x^2y-x^3}\))(1 - \(\dfrac{y-1}{x}\) - \(\dfrac{y}{x^2}\))
18 tháng 2 2022 lúc 20:36
Thực hiện các phép tính sau:
a,(\(\dfrac{x}{x+1}\)+\(\dfrac{x-1}{x}\)):(\(\dfrac{x}{x+1}\)-\(\dfrac{x-1}{x}\))
b,(1+\(\dfrac{x}{y}\)+\(\dfrac{x^2}{y^2}\)).(1-\(\dfrac{x}{y}\)).\(\dfrac{y^2}{x^3-y^3}\)
\(\left(\dfrac{x}{x+1}+\dfrac{x-1}{x}\right):\left(\dfrac{x}{x+1}-\dfrac{x-1}{x}\right)\) \(\left(đk:x\ne0;-1\right)\)
\(=\dfrac{x^2+\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}:\left(\dfrac{x^2-\left(x-1\right)\left(x+1\right)}{x\left(x+1\right)}\right)\)
\(=\dfrac{x^2+x^2-1}{x\left(x+1\right)}.\dfrac{x\left(x+1\right)}{x^2-x^2+1}\)
\(=\dfrac{\left(2x^2-1\right)x\left(x+1\right)}{x\left(x+1\right)}=2x^2-1\)
Đúng 1
Bình luận (0)
Tìm y
a) 3 \(\dfrac{1}{5}\) : 2\(\dfrac{1}{3}\): y = \(\dfrac{12}{7}\)
b) 3 : y x 3 \(\dfrac{1}{2}\)= \(\dfrac{2}{3}x\dfrac{3}{4}\)
c) \(3\dfrac{2}{3}-y+1\dfrac{3}{4}=2\)
mình ko chép đề bài nha
a) \(\dfrac{16}{5}\): \(\dfrac{7}{3}\) : y =\(\dfrac{12}{7}\)
\(\dfrac{48}{35}\): y = \(\dfrac{12}{7}\)
y = \(\dfrac{48}{35}\): \(\dfrac{12}{7}\)
y = \(\dfrac{4}{5}\)
Đúng 2
Bình luận (0)
b) 3 : y x \(\dfrac{7}{2}\)= \(\dfrac{1}{2}\)
3 : y = \(\dfrac{1}{2}:\dfrac{7}{2}\)
3 : y = \(\dfrac{1}{7}\)
y = 3 : \(\dfrac{1}{7}\)
y = 21
Đúng 1
Bình luận (0)
Xem thêm câu trả lời
Rút gọn biểu thức:A [(32)^{dfrac{2}{3}}]^{dfrac{-2}{5}}B dfrac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}C (a^{dfrac{1}{3}}-b^{dfrac{2}{3}})(a^{dfrac{2}{3}}+a^{dfrac{1}{3}}×b^{dfrac{4}{3}}+b^{dfrac{4}{9}})D (x+y^dfrac{3}{2}÷sqrt{x})^dfrac{2}{3}÷[dfrac{sqrt{x}-sqrt{y}}{sqrt{x}}+dfrac{sqrt{y}}{sqrt{x}-sqrt{y}}]^dfrac{2}{3}E [dfrac{1}{x^{dfrac{1}{2}}-4x^{dfrac{-1}{2}}}-dfrac{2sqrt[3]{x}}{xsqrt[3]{x}-4sqrt[3]{x}}]^{-2}-sqrt{x^2+8x+16}
Đọc tiếp
Rút gọn biểu thức:
A = \([(32)^{\dfrac{2}{3}}]^{\dfrac{-2}{5}}\)
B= \(\dfrac{x^{-2}+y^{-2}}{x^{-1}+y^{-1}}\)
C = \((a^{\dfrac{1}{3}}-b^{\dfrac{2}{3}})(a^{\dfrac{2}{3}}+a^{\dfrac{1}{3}}×b^{\dfrac{4}{3}}+b^{\dfrac{4}{9}})\)
D = \((x+y^\dfrac{3}{2}÷\sqrt{x})^\dfrac{2}{3}÷[\dfrac{\sqrt{x}-\sqrt{y}}{\sqrt{x}}+\dfrac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}]^\dfrac{2}{3}\)
E = \([\dfrac{1}{x^{\dfrac{1}{2}}-4x^{\dfrac{-1}{2}}}-\dfrac{2\sqrt[3]{x}}{x\sqrt[3]{x}-4\sqrt[3]{x}}]^{-2}-\sqrt{x^2+8x+16}\)