TÍNH
A,|X-3 Y|^2007+|Y+4|^2008
B,(X+Y)^2016+2017.|Y-1|
tinh
a, |x-3y|^2007+|y+4|^2008
b, (x+y)^2016+2017|y-1|
a, |x-3y|^2007+|y+4|^2008
<=>|x-3y|^2007|=0=>|x-3y|=0 =>x-3y=0 (1)
<=>|y+4|^2008=0=>|y+4|=0=>y+4=0 (2)
tu 1,2 => y=-4 =>x=-12
b, <=>(x+y)^2016=0=>x+y=0 (1)
<=>2017|y-1|=0=>|y-1|=0=>y-1=0 (2)
tu 1, 2 =>y=1=>x=-1
Chất là một từ tặng cho riêng dinhkhachoang
|x+5|+(3y-4)^2016=0
(5x-y)^2016+|x^2-4|^2017<=0
(2x-1)^2014+(y-2/5)^2016+|x+y+z|=0
|x-1|+|x-2|+|y-3|+|x-4|=3
\(\dfrac{\sqrt{x-2015}-1}{x-2015}\) + \(\dfrac{\sqrt{y-2016}-1}{y-2016}\) + \(\dfrac{\sqrt{z-2017}-1}{z-2017}\) = \(\dfrac{3}{4}\)
Đặt \(a=\sqrt{x-2015};b=\sqrt{y-2016};c=\sqrt{z-2017}\left(a,b,c>0\right)\)
Khi đó phương trình trở thành:
\(\dfrac{a-1}{a^2}+\dfrac{b-1}{b^2}+\dfrac{c-1}{c^2}=\dfrac{3}{4}\\ \Leftrightarrow\left(\dfrac{1}{4}-\dfrac{1}{a}+\dfrac{1}{a^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{b}+\dfrac{1}{b^2}\right)+\left(\dfrac{1}{4}-\dfrac{1}{c}+\dfrac{1}{c^2}\right)=0\\ \Leftrightarrow\left(\dfrac{1}{2}-\dfrac{1}{a}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{b}\right)^2+\left(\dfrac{1}{2}-\dfrac{1}{c}\right)^2=0\\ \Leftrightarrow a=b=c=2\\ \Leftrightarrow x=2019;y=2020;z=2021\)
Tick plz
Tìm x ; y biết: \(\hept{\begin{cases}x^{2017}+y^{2017}=1\\\sqrt[2017]{x}-\sqrt[2017]{y}=\left(\sqrt[2016]{y}-\sqrt[2016]{x}\right)\left(x+y+xy+2017\right)\end{cases}}\)
\(\hept{\begin{cases}x^{2017}+y^{2017}=1\left(1\right)\\\sqrt[2017]{x}-\sqrt[2017]{y}=\left(\sqrt[2016]{y}-\sqrt[2016]{x}\right)\left(x+y+xy+2017\right)\left(2\right)\end{cases}}\)
Điều kiện: \(x,y\ge0\)
Dễ thấy \(\hept{\begin{cases}x=0\\y=0\end{cases}}\)không phải là nghiệm của hệ
Đặt \(\hept{\begin{cases}\sqrt[2017.2016]{x}=a>0\\\sqrt[2017.2016]{y}=b>0\end{cases}}\)
\(\Rightarrow\left(2\right)\Leftrightarrow a^{2016}-b^{2016}=\left(b^{2017}-a^{2017}\right)A\left(x,y\right)\)
\(\Leftrightarrow\left(a-b\right).B\left(a,b\right)=\left(b-a\right).C\left(a,b\right).A\left(x,y\right)\)
\(\Leftrightarrow\left(a-b\right)\left(B\left(a,b\right)+C\left(a,b\right).A\left(x,y\right)\right)=0\)
Dễ thấy \(\left(B\left(a,b\right)+C\left(a,b\right).A\left(x,y\right)\right)>0\)
\(\Leftrightarrow a=b\)
\(\Rightarrow\sqrt[2016.2017]{x}=\sqrt[2016.2017]{y}\)
\(\Leftrightarrow x=y\)
Thế vô (1) ta được:
\(2x^{2017}=1\)
\(\Rightarrow x=y=\sqrt[2017]{\frac{1}{2}}\)
cho x,y,z la so thuc thoa man y+z+3/x=x+z+2/y=x+y-3/z=1/x+y+z.
Tinh A=2016.x+y^2017+z^2017
Câu hỏi của Phung Thi Thanh Thao - Toán lớp 7 - Học toán với OnlineMath
Tham khảo tính được x,y,z.Thay vào A
Cho x,y>0 thỏa mãn
x^2015+y^2015=x^2016+y^2016=x^2017+y^2017
C/m: 1/x^2018+1/y^2018=1/x^2019+1/y^2019
Giải pt
1)x+y+z+8=\(2\sqrt{x-1}\)+\(4\sqrt{y-2}\)+\(6\sqrt{z-3}\)
2)\(\sqrt{x}+\sqrt{x+1}=1\)
3)\(\left(1+\sqrt{x^2+2017+2016}\right)\)\(\left(\sqrt{2016+x}-\sqrt{x+1}\right)\)=2015
1.
ĐKXĐ: $x\geq 1; y\geq 2; z\geq 3$
PT \(\Leftrightarrow x+y+z+8-2\sqrt{x-1}-4\sqrt{y-2}-6\sqrt{z-3}=0\)
\(\Leftrightarrow [(x-1)-2\sqrt{x-1}+1]+[(y-2)-4\sqrt{y-2}+4]+[(z-3)-6\sqrt{z-3}+9]=0\)
\(\Leftrightarrow (\sqrt{x-1}-1)^2+(\sqrt{y-2}-2)^2+(\sqrt{z-3}-3)^2=0\)
\(\Rightarrow \sqrt{x-1}-1=\sqrt{y-2}-2=\sqrt{z-3}-3=0\)
\(\Leftrightarrow \left\{\begin{matrix} x=2\\ y=6\\ z=12\end{matrix}\right.\)
2.
ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow \sqrt{x+1}=1-\sqrt{x}$
$\Rightarrow x+1=(1-\sqrt{x})^2=x+1-2\sqrt{x}$
$\Leftrightarrow 2\sqrt{x}=0$
$\Leftrightarrow x=0$
Thử lại thấy thỏa mãn
Vậy $x=0$
3.
ĐKXĐ: $x\geq -1$
PT \(\Leftrightarrow (1+\sqrt{x^2+4033}).\frac{(x+2016)-(x+1)}{\sqrt{x+2016}+\sqrt{x+1}}=2015\)
\(\Leftrightarrow 1+\sqrt{x^2+4033}=\sqrt{x+2016}+\sqrt{x+1}\)
\(\Leftrightarrow (1+\sqrt{x^2+4033})^2=(\sqrt{x+2016}+\sqrt{x+1})^2\)
Áp dụng BĐT Bunhiacopxky:
\(\text{VP}\leq 2(x+2016+x+1)=4x+4034\)
\(\text{VP}=x^2+4034+2\sqrt{x^2+4033}\geq x^2+4034+2\sqrt{4033}>x^2+4034+5\)
Mà: $x^2+4034+5-(4x+4034)=(x-2)^2+1> 0$
$\Rightarrow x^2+4034+5> 4x+4034$
$\Rightarrow \text{VP}> \text{VT}$
Do đó pt vô nghiệm.
Cho hai số x,y thỏa mãn (x-2)^2016 + |y+1| = 0. Tính giá trị biểu thức A= 2.x^2.y^2016 - 3.(x+ y)^2017
tìm x,y biết x^2015 +x^2016+2015^2016=y^2016+y^2017+2016^2017