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\(a,A=\dfrac{\dfrac{3}{4}-\dfrac{3}{11}+\dfrac{3}{13}}{\dfrac{5}{7}-\dfrac{5}{11}+\dfrac{5}{13}}+\dfrac{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}}{\dfrac{5}{4}-\dfrac{5}{6}+\dfrac{5}{8}}\\ A=\dfrac{\dfrac{405}{572}}{\dfrac{645}{1001}}+\dfrac{\dfrac{5}{12}}{\dfrac{25}{24}}\\ A=\dfrac{189}{172}+\dfrac{2}{5}\\ A=\dfrac{1289}{860}\)
Câu 1: Mình chỉnh sửa lại đầu bài của bạn nha. Không biết có đúng không. Nếu để đầu bài như bạn thì mình không làm ra được. Mog góp ý !!!!
Áp dụng t/c DTSBN ta có:
\(\dfrac{x}{y+z+1}=\dfrac{y}{x+z+1}=\dfrac{z}{x+y-2}=x+y+z\)
\(=\dfrac{x+y+x}{y+z+1+x+z+1+x+y-2}=\dfrac{x+y+x}{2x+2y+2z}=\dfrac{1}{2}\)
=>\(\dfrac{x}{y+z+1}=\dfrac{1}{2}\left(1\right)\)
=>\(\dfrac{y}{x+z+1}=\dfrac{1}{2}\left(2\right)\)
=>\(\dfrac{z}{x+y-2}=\dfrac{1}{2}\left(3\right)\)
=> x+y+z = 1/2 (4)
Ta có : Từ (1) => 2x = y+z+1 kết hợp (4)
=> 2x = 1/2-x+1
=> 3x = 3/2 => x=1/2
Ta có: Từ (2) => 2y = x+z+1
=> 2y + y = x+y+z+1
=> 3y = 1/2+1 (theo 4) => 3y=3/2
=> y=1/2
Ta có : Từ (4) => x+y+z=1/2
=>1/2 + 1/2 +z = 1/2
=> z=-1/2
Vậy ( x;y;z)=(1/2;1/2;-1/2)
b: 2x^3-1=15
=>2x^3=16
=>x=2
\(\dfrac{x+16}{9}=\dfrac{y-25}{16}=\dfrac{z+9}{25}\)
=>\(\dfrac{y-25}{16}=\dfrac{z+9}{25}=\dfrac{18}{9}=2\)
=>y-25=32; z+9=50
=>y=57; z=41
d: 3/5x=2/3y
=>9x=10y
=>x/10=y/9=k
=>x=10k; y=9k
x^2-y^2=38
=>100k^2-81k^2=38
=>19k^2=38
=>k^2=2
TH1: k=căn 2
=>\(x=10\sqrt{2};y=9\sqrt{2}\)
TH2: k=-căn 2
=>\(x=-10\sqrt{2};y=-9\sqrt{2}\)
a, H = \(2^{2010}-2^{2009}-2^{2008}-...-2-1\)
\(\Leftrightarrow\) 2H = \(2^{2011}-2^{2010}-2^{2009}-...-2^2-2\)
\(\Leftrightarrow\) 2H - H = \((2^{2011}-2^{2010}-2^{2009}-...-2^2-2)\) - \((2^{2010}-2^{2009}-2^{2008}-...-2-1)\)
\(\Leftrightarrow\) H = \(2^{2011}-2.2^{2010}+1\)
\(\Leftrightarrow\) H = \(2^{2011}-2^{2011}+1\)
\(\Leftrightarrow\) H = 1
Vậy H = 1
a)H=22010-22009-...-2-1
=>2H=2(22010-22009-...-2-1)
=>2H=22011-22010-...-22-2
=>2H-H=(22011-22010-...-22-2)-(22010-22009-...-2-1)
=>H=22011-1
Ta có:
\(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\)
\(\Rightarrow\dfrac{x}{y+z+t}+1=\dfrac{y}{z+t+x}+1=\dfrac{z}{t+x+y}+1=\dfrac{t}{x+y+z}+1\)
\(\Rightarrow\dfrac{x+y+z+t}{y+z+t}=\dfrac{x+y+z+t}{z+t+x}=\dfrac{x+y+z+t}{y+x+x}=\dfrac{x+y+z+t}{y+x+z}\)
. Xét TH1: \(x+y+z+t=0\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=-\left(z+t\right)\\y+z=-\left(x+t\right)\\z+t=-\left(x+y\right)\\x+t=-\left(y+z\right)\end{matrix}\right.\)
. Xét TH2: \(x+y+z+t\ne0\)
\(\Rightarrow x=y=z=t\)
\(\Rightarrow A=1\)
\(\Rightarrow\left\{{}\begin{matrix}A=1\\A=-1\end{matrix}\right.\)
ta có :\(\dfrac{y+z-2015x}{x}=\dfrac{z+x-2015y}{y}=\dfrac{z+y-2015z}{z}\)
=>\(\left(\dfrac{y+z-2015}{x}+2016\right)=\left(\dfrac{z+x-2015y}{y}+2016\right)=\left(\dfrac{x+y-2015z}{z}+2016\right)\)
(=)\(\dfrac{x+y+z}{x}=\dfrac{x+y+z}{y}=\dfrac{x+y+z}{z}\)
*Nếu x+y+z\(\ne\)0
=>\(\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)
=>\(P=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)=1.1.1=1
*Nếu x+y+z=0
=>x=y=z
=>\(P=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)=2.2.2=8
@ Mashiro Shiina
@Akai Haruma
@Nguyễn Thanh Hằng
@Đẹp Trai Không Bao Giờ Sai
Có: \(\dfrac{y+z-x}{x}=\dfrac{x+z-y}{y}=\dfrac{x+y-z}{z}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{y+z-x}{x}=\dfrac{x+z-y}{y}=\dfrac{x+y-z}{z}=\dfrac{x+y+z}{x+y+z}=1\)
Vì
\(\dfrac{y+z-x}{x}=\dfrac{z+x-y}{y}=\dfrac{x+y+z}{z}\)
\(\Rightarrow\dfrac{y+z-x}{x}+2=\dfrac{z+x-y}{y}+2=\dfrac{x+y-z}{z}+2=\)
\(\dfrac{y+z+x}{x}=\dfrac{z+x+y}{y}=\dfrac{x+y+z}{z}\)
\(\Rightarrow\)x=y=z\(\Rightarrow\)\(\dfrac{x}{y}=\dfrac{y}{z}=\dfrac{z}{x}=1\)
\(\Rightarrow\)B=(1+1)(1+1)(1+1)=8