1. Cho \(x,y\) thỏa mãn \(\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\)
Tính \(x+y\)
2. Cho \(a,b\ne-2\) thỏa mãn \(\left(2a+1\right)\left(2b+1\right)=9\)
Tính \(A=\dfrac{1}{2+a}+\dfrac{1}{2+b}\)
Cho 2 số thực x, y thỏa mãn:
\(\left(x+\sqrt{x^2+2020}\right)\left(2y+\sqrt{4y^2+2020}\right)=2020\)
Tìm GTLN cuẩ biểu thức: B=\(\dfrac{x^2}{2}+4xy+3y^2+x+3y+15\)
\(\left(x+\sqrt{x^2+2020}\right)\left(2y+\sqrt{\left(2y\right)^2+2020}\right)=2020\)
\(\Leftrightarrow\left\{{}\begin{matrix}2y+\sqrt{\left(2y\right)^2+2020}=\sqrt{x^2+2020}-x\\x+\sqrt{x^2+2020}=\sqrt{\left(2y\right)^2+2020}-2y\end{matrix}\right.\)
\(\Rightarrow x+2y+\sqrt{x^2+2020}+\sqrt{\left(2y\right)^2+2020}=-x-2y+\sqrt{x^2+2020}+\sqrt{\left(2y\right)^2+2020}\)
\(\Leftrightarrow2\left(x+2y\right)=0\)
\(\Leftrightarrow x=-2y\)
\(\Rightarrow B=2y^2-8y^2+3y^2-2y+3y+15\)
\(\Rightarrow B=-3y^2+y+15=-3\left(y-\dfrac{1}{6}\right)^2+\dfrac{181}{12}\)
\(B_{max}=\dfrac{181}{12}\) khi \(y=\dfrac{1}{6}\)
1. Cho \(\left(x\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\)
Tính S=x+y+2020
`(x+sqrt{x^2+2020})(sqrt{x^2+2020}-x)=x^2+2020-x^2=2020`
`=>y+sqrt{y^2+2020}=sqrt{x^2+2020}-x`
`<=>x+y=sqrt{x^2+2020}-sqrt{y^2+2020}`
Tương tự:`x+y=sqrt{y^2+2020}-sqrt{x^2+2020}`
Cộng từng vế
`=>2(x+y)=0`
`<=>S=0+2020=2020`
Gt\(\Leftrightarrow\left(x+\sqrt{x^2+2020}\right)\left(x-\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\left(x-\sqrt{x^2+2020}\right)\)
\(\Leftrightarrow\left(x^2-x^2-2020\right)\left(y+\sqrt{y^2+2020}\right)=2020\left(x-\sqrt{x^2+2020}\right)\)
\(\Leftrightarrow-y-\sqrt{y^2+2020}=x-\sqrt{x^2+2020}\) (1)
Gt\(\Leftrightarrow\left(x+\sqrt{x^2+2020}\right)\left(y-\sqrt{y^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=2020\left(y-\sqrt{y^2+2020}\right)\)
\(\Leftrightarrow\left(y^2-y^2-2020\right)\left(x+\sqrt{x^2+2020}\right)=2020\left(y-\sqrt{y^2+2020}\right)\)
\(\Leftrightarrow-x-\sqrt{x^2+2020}=y-\sqrt{y^2+2020}\) (2)
Từ (1) (2) cộng vế với vế \(\Rightarrow-\left(x+y\right)-\left(\sqrt{y^2+2020}+\sqrt{x^2+2020}\right)=x+y-\left(\sqrt{y^2+2020}+\sqrt{x^2+2020}\right)\)
\(\Leftrightarrow-2\left(x+y\right)=0\)
\(\Leftrightarrow x+y=0\)
\(S=x+y+2020=2020\)
Cho các số thực dương \(x,y,z\) thỏa mãn: \(xy+yz+xz=1\). Hãy tính giá trị biểu thức: \(A=x\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}+y\sqrt{\dfrac{\left(1+z^2\right)\left(1+x^2\right)}{1+y^2}}+z\sqrt{\dfrac{\left(1+x^2\right)\left(1+y^2\right)}{1+z^2}}\)
Ta có:
\(x^2+1=x^2+xy+yz+zx\)
\(=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)
Tương tự:
\(\left\{{}\begin{matrix}y^2+1=\left(y+z\right)\left(y+x\right)\\z^2+1=\left(z+y\right)\left(z+x\right)\end{matrix}\right.\)
\(A=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\dfrac{\left(z+x\right)\left(y+z\right)\left(x+y\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)
\(=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)
TH1: x,y,z <0
\(A=-x\left(y+z\right)-y\left(z+x\right)-z\left(x+y\right)=-2\)
TH2: x,y,z>0
\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\)
Ta có \(1+z^2=xy+yz+zx+z^2\)
\(=y\left(x+z\right)+z\left(x+z\right)\)
\(=\left(x+z\right)\left(y+z\right)\)
CMTT, \(1+x^2=\left(x+y\right)\left(x+z\right)\) và \(1+y^2=\left(x+y\right)\left(y+z\right)\)
Do đó \(\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\) \(=\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)
\(=\sqrt{\left(y+z\right)^2}\) \(=\left|y+z\right|\)
Tương tự như thế, ta được
\(A=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)
Cái này không tính ra số cụ thể được nhé bạn. Nó còn phải tùy vào dấu của \(x+y,y+z,z+x\) nữa.
Cho các số x,y thỏa mãn điều kiện:
\(x^2-2xy+6y^2-12x+2y+41=0\)
Tính giá trị của biểu thức: A=\(\dfrac{2020-2019\left(9-x-y\right)^{2019}-\left(x-6y\right)^{2010}}{y^{2010}}\)
Cho ba số x,y,z thỏa mãn: \(\dfrac{x}{2018}=\dfrac{y}{2019}=\dfrac{z}{2020}\)
CMR: \(\left(x-z\right)^3=8\left(x-y\right)^2\left(y-z\right)\)
\(\dfrac{x}{2018}=\dfrac{y}{2019}=\dfrac{x-y}{-1};\dfrac{y}{2019}=\dfrac{z}{2020}=\dfrac{y-z}{-1};\dfrac{x}{2018}=\dfrac{z}{2020}=\dfrac{x-z}{-2}\\ \Leftrightarrow\dfrac{x-y}{-1}=\dfrac{y-z}{-1}=\dfrac{x-z}{-2}\\ \Leftrightarrow2\left(x-y\right)=2\left(y-z\right)=x-z\\ \Leftrightarrow\left(x-z\right)^3=8\left(x-y\right)^3=8\left(x-y\right)^2\left(x-y\right)=8\left(x-y\right)^2\left(y-z\right)\)
Bài 1: Cho a, b thỏa mãn ab > 2020a + 2021b
Chứng minh rằng: a+b > \(\left(\sqrt{2020}+\sqrt{2021}\right)^2\)
Bài 2: Tìm x,y thỏa mãn \(\sqrt{x-3}+\sqrt{5-x}=y^2+2\sqrt{2019}.y+2021\)
bài 1 ta có
\(\left(\frac{1}{a}+\frac{1}{b}\right)\left(2020a+2021b\right)\ge\left(\sqrt{2020}+\sqrt{2021}\right)^2\) ( BDT Bunhia )
do đó
\(a+b=ab.\left(\frac{1}{a}+\frac{1}{b}\right)\ge\left(\frac{1}{a}+\frac{1}{b}\right)\left(2020a+2021b\right)\ge\left(\sqrt{2020}+\sqrt{2021}\right)^2\)
vậy ta có đpcm.
bài 2.
ta có \(VT=\sqrt{x-3}+\sqrt{5-x}\le2\)( BDT Bunhia )
\(VP=y^2+2.\sqrt{2019}y+2021=\left(y+\sqrt{2019}\right)^2+2\ge2\)
suy ra PT có nghiệm \(\hept{\begin{cases}x-3=5-x\\y+\sqrt{2019}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=4\\y=-\sqrt{2019}\end{cases}}}\)
Cho x, y, z dương thỏa mãn xyz=1. Tìm GTLN của \(\dfrac{1}{\sqrt{\left(x+y\right)^2+\left(x+1\right)^2+4}}+\dfrac{1}{\sqrt{\left(y+z\right)^2+\left(y+1\right)^2+4}}+\dfrac{1}{\sqrt{\left(z+x\right)^2+\left(z+1\right)^2+4}}\)
\(P\le\sqrt{3\left(\sum\dfrac{1}{\left(x+y\right)^2+\left(x+1\right)^2+4}\right)}\le\sqrt{3\left(\sum\dfrac{1}{4xy+4x+4}\right)}\)
\(P\le\sqrt{\dfrac{3}{4}\sum\left(\dfrac{1}{xy+x+1}\right)}=\dfrac{\sqrt{3}}{2}\)
\(P_{max}=\dfrac{\sqrt{3}}{2}\) khi \(x=y=z=1\)
Cho cách số thực x, y thỏa mãn xy > 2020x+2020yChứng minh x + y > \(\left(\sqrt{2020}+\sqrt{2021}\right)^2\)
Đề sai. Nếu $x,y$ đều âm thì điều kiện $xy> 2020x+2020y$ được thỏa mãn nhưng hiển nhiên $x+y$ không thể lớn hơn $(\sqrt{2020}+\sqrt{2021})^2$
Cho 3 số x, y, z thỏa mãn: \(\dfrac{x}{2018}=\dfrac{y}{2019}=\dfrac{z}{2020}\)
CMR: \(\left(x-z\right)^3=8\left(x-y\right)^2\left(y-z\right)\)
HELP ME!
Lời giải:
Đặt $\frac{x}{2018}=\frac{y}{2019}=\frac{z}{2020}=a$
$\Rightarrow x=2018a; y=2019a; z=2020a$
$\Rightarrow (x-z)^3=(2018a-2020a)^3=(-2a)^3=-8a^3(1)$
Mặt khác:
$8(x-y)^2(y-z)=8(2018a-2019a)^2(2019a-2020a)=8a^2.(-a)=-8a^3(2)$
Từ $(1); (2)$ ta có đpcm.