Cho B=\(\dfrac{x\sqrt{x}-2\sqrt{x}}{x\left(\sqrt{x}+1\right)}\)
C/m: \(\dfrac{B^{2021}+1}{B^{2020}+1}>B\)
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 \(A=\dfrac{1}{2+a}+\dfrac{1}{2+b}\)
Bài 1.
Ta có:\(\left(x+\sqrt{x^2+2020}\right)\left(\sqrt{x^2+2020}-x\right)=x^2+2020-x^2=2020\)
\(\Rightarrow\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=\left(x+\sqrt{x^2+2020}\right)\left(\sqrt{x^2+2020}-x\right)\)
\(\Rightarrow y+\sqrt{y^2+2020}=\sqrt{x^2+2020}-x\)
\(\Rightarrow x+y=\sqrt{x^2+2020}-\sqrt{y^2+2020}\) (1)
Ta có:\(\left(y+\sqrt{y^2+2020}\right)\left(\sqrt{y^2+2020}-y\right)=y^2+2020-y^2=2020\)
\(\Rightarrow\left(x+\sqrt{x^2+2020}\right)\left(y+\sqrt{y^2+2020}\right)=\left(y+\sqrt{y^2+2020}\right)\left(\sqrt{y^2+2020}-y\right)\)
\(\Rightarrow x+\sqrt{x^2+2020}=\sqrt{y^2+2020}-y\)
\(\Rightarrow x+y=\sqrt{y^2+2020}-\sqrt{x^2+2020}\) (2)
Cộng vế với vế của (1) và (2) ta có:
\(2\left(x+y\right)=\sqrt{y^2+2020}-\sqrt{x^2+2020}+\sqrt{x^2+2020}-\sqrt{y^2+2020}\)
\(\Rightarrow2\left(x+y\right)=0\Rightarrow x+y=0\)
Bài 2:
Ta có: (2a+1)(2b+1)=9
nên \(2b+1=\dfrac{9}{2a+1}\)
\(\Leftrightarrow2b=\dfrac{9}{2a+1}-\dfrac{2a+1}{2a+1}=\dfrac{8-2a}{2a+1}\)
\(\Leftrightarrow b=\dfrac{8-2a}{4a+2}=\dfrac{4-a}{2a+1}\)
\(\Leftrightarrow b+2=\dfrac{4-a+4a+2}{2a+1}=\dfrac{3a+6}{2a+1}\)
Ta có: \(A=\dfrac{1}{a+2}+\dfrac{1}{b+2}\)
\(=\dfrac{1}{a+2}+\dfrac{2a+1}{3a+6}\)
\(=\dfrac{3+2a+1}{3a+6}\)
\(=\dfrac{2a+4}{3a+6}=\dfrac{2}{3}\)
(1,0 điểm) 1. Giải phương trình $4x+1-\sqrt{9\left( 2x-1 \right)\left( x+1 \right)}+2\sqrt{2x-1}-2\sqrt{x+1}=0$ (1).
2. Cho các số thực dương $a,$ $b,$ $c$ thỏa mãn $a+b+c=3$. Tìm giá trị nhỏ nhất của biểu thức $P=\dfrac{2021}{\sqrt{ab}+\sqrt{cd}+\sqrt{ac}}-\dfrac{b\sqrt{a}}{1+b}-\dfrac{c\sqrt{b}}{1+c}-\dfrac{a\sqrt{c}}{1+a}$.
Cho biểu thức B =\(\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)
a) Tìm điều kiện để B có nghĩa
b) Rút gọn B
c) Tính B với x =\(\dfrac{2-\sqrt{3}}{2}\)
a) ĐKXĐ : \(x\sqrt{x}-1\ge0\Leftrightarrow x\ge1\)
b) \(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right).\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\dfrac{2x+1-\sqrt{x}.\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)}.\left(x-2\sqrt{x}+1\right)\)
\(=\dfrac{1}{\sqrt{x}-1}.\left(\sqrt{x}-1\right)^2=\sqrt{x}-1\)
c) Có : \(x=\dfrac{2-\sqrt{3}}{2}=\dfrac{4-2\sqrt{3}}{4}=\dfrac{\left(\sqrt{3}-1\right)^2}{4}\)
Khi đó B = \(\dfrac{\sqrt{3}-1}{2}-1=\dfrac{\sqrt{3}-3}{2}\)
\(a,\) B có nghĩa \(\Leftrightarrow\left[{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
\(b,B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(=\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{1+x\sqrt{x}-\sqrt{x}\left(1+\sqrt{x}\right)}{1+\sqrt{x}}\)
\(=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{1+x\sqrt{x}-\sqrt{x}-x}{1+\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{\sqrt{x}\left(x-1\right)-\left(x-1\right)}{1+\sqrt{x}}\)
\(=\dfrac{\left(x-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\sqrt{x}-1\)
\(c,x=\dfrac{2-\sqrt{3}}{2}\Rightarrow B=\sqrt{\dfrac{2-\sqrt{3}}{2}}-1\)
\(=\dfrac{\sqrt{2}.\sqrt{2-\sqrt{3}}}{\sqrt{2}.\sqrt{2}}-\sqrt{2}\) (Nhân \(\sqrt{2}\) để khử căn dưới mẫu)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-2\sqrt{2}}{2}\)
\(=\dfrac{\sqrt{\left(\sqrt{3}-1\right)^2}-2\sqrt{2}}{2}\)
\(=\dfrac{\left|\sqrt{3}-1\right|-2\sqrt{2}}{2}\)
\(=\dfrac{\sqrt{3}-1-2\sqrt{2}}{2}\)
Cho biểu thức
1) A=\(\dfrac{\sqrt{x}+2}{2\sqrt{x}-4}+\dfrac{\sqrt{x}-2}{2\sqrt{x}+4}\)
a. Rút gọn A
b. Tìm x để A=8
2) C= \(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
a. Rút gọn C
b.Tìm x để C= -8
3) D=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\dfrac{\sqrt{4x}}{x-4}\)
a.Rút gọn D
b.Tìm x để D>3
Bài 1:
a: \(A=\dfrac{\sqrt{x}+2}{2\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{2\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+4\sqrt{x}+4+x-4\sqrt{x}+4}{2\left(x-4\right)}\)
\(=\dfrac{2x+8}{2\left(x-4\right)}=\dfrac{x+4}{x-4}\)
b: Để A=8 thì x+4=8(x-4)
=>x+4=8x-32
=>-7x=-36
hay x=36/7(nhận)
tìm các khoảng đơn điệu của hàm số
a)y = \(\dfrac{-2x+1}{x^2-3x+1}\)
b)y = \(x\left(2021+\sqrt{2020-x^2}\right)\)
Cho B=\(\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}-\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
a)Rút gọn B
b)Tìm m để với mọi giá trị x>9 ta có \(m\left(\sqrt{x}-3\right)B>x+1\)
a: \(=\dfrac{4x-8\sqrt{x}+8x}{x-4}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4\sqrt{x}\left(3\sqrt{x}-2\right)}{x-4}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}=\dfrac{-4x\left(3\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\)
b: \(m\left(\sqrt{x}-3\right)\cdot B>x+1\)
=>\(-4xm\left(3\sqrt{x}-2\right)>\left(\sqrt{x}+2\right)\cdot\left(x+1\right)\)
=>\(-12m\cdot x\sqrt{x}+8xm>x\sqrt{x}+2x+\sqrt{x}+2\)
=>\(x\sqrt{x}\left(-12m-1\right)+x\left(8m-2\right)-\sqrt{x}-2>0\)
Để BPT luôn đúng thì m<-0,3
1.A=\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}}{2}\)
2.B=\(\left(\dfrac{2\sqrt{x+x+1}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right):\left(1-\sqrt{x}\right)\)
3.C=\(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\dfrac{8\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}-x-3}{x-1}-\dfrac{1}{\sqrt{x}-1}\right)\)
Làm chi tiết hộ mình với ak mình đang cần gấp!!!
1: Ta có: \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}}{2}\)
\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-2}{\sqrt{x}\left(x+\sqrt{x}+1\right)}\)
Cho biểu thức: B = \(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{\sqrt{x}}{1-x}\right)\) : \(\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{1-\sqrt{x}}{\sqrt{x}+1}\right)\) . Tính gtri của B biết x = \(\dfrac{2-\sqrt{3}}{2}\)
ĐKXĐ: x>=0; x<>1
\(B=\dfrac{\left(\sqrt{x}+1\right)^2+\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}}{x-1}:\dfrac{\left(\sqrt{x}+1\right)^2-\left(\sqrt{x}-1\right)^2}{x-1}\)
\(=\dfrac{x+2\sqrt{x}+1+x-\sqrt{x}+\sqrt{x}}{x-1}\cdot\dfrac{x-1}{x+2\sqrt{x}+1-x+2\sqrt{x}-1}\)
\(=\dfrac{2x+2\sqrt{x}+1}{4\sqrt{x}}\)
Khi \(x=\dfrac{2-\sqrt{3}}{2}=\dfrac{4-2\sqrt{3}}{4}=\left(\dfrac{\sqrt{3}-1}{2}\right)^2\) thì:
\(B=\dfrac{2\cdot\dfrac{2-\sqrt{3}}{2}+2\cdot\dfrac{\sqrt{3}-1}{2}+1}{4\cdot\dfrac{\sqrt{3}-1}{2}}\)
\(=\dfrac{2-\sqrt{3}+\sqrt{3}-1+1}{2\left(\sqrt{3}-1\right)}=\dfrac{2}{2\left(\sqrt{3}-1\right)}=\dfrac{1}{\sqrt{3}-1}=\dfrac{\sqrt{3}+1}{2}\)
1. Cho biểu thức : B = \(\left(\dfrac{\sqrt{x}}{2}-\dfrac{1}{2\sqrt{x}}\right)^{^2}\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right).\)
a) Rút gọn B.
b) Tìm x để B \(< 0\).
c) Tìm x để B = 2.
2. Tìm GTLN của A = \(\sqrt{1-x}+\sqrt{1+x}.\)
Bài 1:
a: \(A=\left(\dfrac{x-1}{2\sqrt{x}}\right)^2\cdot\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}\)
\(=\dfrac{\left(x-1\right)^2}{4x}\cdot\dfrac{-4\sqrt{x}}{x-1}=\dfrac{-\left(x-1\right)}{\sqrt{x}}\)
b: Để B<0 thì -x+1<0
=>-x<-1
hay x>1
c: Để B=2 thì \(-\left(x-1\right)=2\sqrt{x}\)
\(\Leftrightarrow-x+1-2\sqrt{x}=0\)
\(\Leftrightarrow x+\sqrt{x}-1=0\)
\(\Leftrightarrow\sqrt{x}=\dfrac{\sqrt{5}-1}{2}\)
hay \(x=\dfrac{6-2\sqrt{5}}{4}\)