cho bt
A=\(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}\)-\(\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\)+\(\dfrac{x+1}{\sqrt{x}}\)
a, rút gọn P
b, tìm x để A=\(\dfrac{1}{2}\)
c, chứng minh rằng A< 0
cho bt A=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\dfrac{3\sqrt{x}+1}{x-1}\)(với x≥0;x≠1)
1)rút gọn bt A
2)tính A khi x=9
3)tìm x để A <1
1) \(A=\dfrac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(2x-2\sqrt{x}\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b) \(A=\dfrac{2\sqrt{9}-1}{\sqrt{9}+1}=\dfrac{5}{4}\)
c) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}< 1\Rightarrow2\sqrt{x}-1< \sqrt{x}+1\Rightarrow\sqrt{x}< 2\Rightarrow x< 4\)
\(1,A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ A=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\\ 2,x=9\Leftrightarrow A=\dfrac{6-1}{3+1}=\dfrac{5}{4}\\ 3,A< 1\Leftrightarrow\dfrac{2\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}+1}< 0\\ \Leftrightarrow\dfrac{\sqrt{x}-2}{\sqrt{x}+1}< 0\Leftrightarrow\sqrt{x}-2< 0\left(\sqrt{x}+1>0\right)\\ \Leftrightarrow x< 4\Leftrightarrow0\le x< 4\)
a) tính gtrị của biểu thức A = \(\sqrt{3}+\sqrt{12}-\sqrt{27}-\sqrt{36}\)
b) cho bt B = \(\dfrac{2}{\sqrt{x-1}}-\dfrac{1}{\sqrt{x}}+\dfrac{3\sqrt{x-5}}{\sqrt{x\left(\sqrt{x-1}\right)}}\) với x > 0 và x \(\ne\) 1 . rút gọn bt và tìm x để B = 2
\(a,\) ta có :
\(\Leftrightarrow\left\{{}\begin{matrix}A=\sqrt{3}+\sqrt{2^2.3}-\sqrt{3^2.3}-\sqrt{6^2}\\A=\sqrt{3}+2\sqrt{3}-3\sqrt{3}-6\\A=\sqrt{3}.\left(1+2-3\right)-6\\A=-6\end{matrix}\right.\)
\(\Rightarrow A=-6\) . vậy \(A=9\sqrt{5}\)
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\(b,\) với \(x>0\) và \(x\ne1\) . ta có :
\(B=\dfrac{2}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}+\dfrac{3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{x}-\left(\sqrt{x}-1\right)+3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{2\sqrt{x}-\sqrt{x}+1+3\sqrt{x}-5}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\) \(B=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow B=\dfrac{4}{\sqrt{x}}\)
vậy với \(x>0\) \(;\) \(x\ne1\) thì \(B=\dfrac{4}{\sqrt{x}}\)
để \(B=2\) thì \(\dfrac{4}{\sqrt{x}}=2\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)
vậy để \(B=2\) thì \(x=4\)
Cho biểu thức: A = \(\dfrac{x-1}{\sqrt{x}}\) : (\(\dfrac{\sqrt{x}-1}{\sqrt{x}}\) + \(\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\)) với x > 0; x ≠ 1
1) Rút gọn A
2) Tìm x để A . \(\sqrt{x}\)= 25
3) Chứng minh A > 4
1) \(A=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{x+\sqrt{x}}\right)\)
\(A=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{\sqrt{x}-1}{\sqrt{x}}+\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(A=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(A=\dfrac{x-1}{\sqrt{x}}:\left(\dfrac{x-1+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
\(A=\dfrac{x-1}{\sqrt{x}}:\dfrac{x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-\sqrt{x}}\)
\(A=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}\cdot\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
b) Ta có:
\(A\cdot\sqrt{x}=25\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\cdot\sqrt{x}=25\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=25\)
\(\Leftrightarrow\left(\sqrt{x}+1\right)^2=5^2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=5\\\sqrt{x}+1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=16\\\sqrt{x}=-6\text{(vô lý)}\end{matrix}\right.\)
c) Ta xét hiệu:
\(A-4=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}-4\)
\(A-4=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}-\dfrac{4\sqrt{x}}{\sqrt{x}}\)
\(A-4=\dfrac{x+2\sqrt{x}+1-4\sqrt{x}}{\sqrt{x}}\)
\(A-4=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}}\)
\(A-4=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)
Với \(x>0\) thì \(\left(\sqrt{x}-1\right)>0\) và \(\sqrt{x}>0\)
\(\Rightarrow\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
Nên A > 4 (đpcm)
1: \(A=\dfrac{x-1}{\sqrt{x}}:\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)+1-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x-1+1-\sqrt{x}}\)
\(=\dfrac{\left(x-1\right)\cdot\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
2: A*căn x=25
=>(căn x+1)^2=25
=>căn x+1=5
=>x=16
3: \(A-4=\dfrac{\left(\sqrt{x}+1\right)^2-4\sqrt{x}}{\sqrt{x}}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}>0\)
=>A>4
Bài 1:
A=\(\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
a) Tìm tập xác định của biểu thức A
b) Rút gọn biểu thức A
c) Chứng minh rằng A>0 với mọi x≠1
d) Tìm x để A đạt GTLN, tìm GTLN đó
a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
b: Ta có: \(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{2}{x+\sqrt{x}+1}\)
c: Ta có: \(x+\sqrt{x}+1>0\forall x\) thỏa mãn ĐKXĐ
\(\Leftrightarrow\dfrac{2}{x+\sqrt{x}+1}>0\forall x\)
Cho \(A=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
a ) Rút gọn A
b ) Tính giá trị biểu thức A khi x = \(28-6\sqrt{3}\)
c ) Chứng minh rằng : A < \(\dfrac{1}{3}\)
cho bt A=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left[\dfrac{2}{x}-\dfrac{2-x}{x\left(\sqrt{x}+1\right)}\right]\)
a)rút gọn bt A
b)tính giá trị của bt A khi\(x=4+2\sqrt{3}\)
c)tìm giá trị của x để bt \(\sqrt{A}\)có giá trị nỏ nhất
ĐKXĐ: \(x>0;x\ne1\)
\(A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{2\left(\sqrt{x}+1\right)}{x\left(\sqrt{x}+1\right)}-\dfrac{2-x}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\left(\dfrac{x+2\sqrt{x}}{x\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\left(x+2\sqrt{x}\right).x.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\left(x+2\sqrt{x}\right)}=\dfrac{x}{\sqrt{x}-1}\)
b.
\(x=4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\Rightarrow\sqrt{x}=\sqrt{3}+1\)
\(\Rightarrow A=\dfrac{4+2\sqrt{3}}{\sqrt{3}+1-1}=\dfrac{4+2\sqrt{3}}{\sqrt{3}}=\dfrac{6+4\sqrt{3}}{3}\)
c.
Để \(\sqrt{A}\) xác định \(\Rightarrow\sqrt{x}-1>0\Rightarrow x>1\)
Ta có:
\(\sqrt{A}=\sqrt{\dfrac{x}{\sqrt{x}-1}}=\sqrt{\dfrac{x}{\sqrt{x}-1}-4+4}=\sqrt{\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-1}+4}\ge\sqrt{4}=2\)
Dấu "=" xảy ra khi \(\sqrt{x}-2=0\Rightarrow x=4\)
P=[\(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\)] :\(\dfrac{\sqrt{x}-1}{2}\)
a)Rút gọn biểu thức trên
b)Chứng minh rằng P > 0 với mọi x≥ 0 và x ≠ 1.
a: \(P=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}}{x\sqrt{x}-1}\)
a, Với x ≥ 0, x ≠1
P= [ \(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\)] : \(\dfrac{\sqrt{x}-1}{2}\) =
\(\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)]
: \(\dfrac{\sqrt{x}-1}{2}\)
P= \(\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\):\(\dfrac{\sqrt{x}-1}{2}\)
P= \(\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}:\dfrac{\sqrt{x}-1}{2}\)
P= \(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
P= \(\dfrac{2}{x+\sqrt{x}+1}\)
b, Ta có : \(x+\sqrt{x}+1=\left(\sqrt{x}\right)^2+2.\dfrac{1}{2}.\sqrt{x}+\dfrac{1}{4}+\dfrac{3}{4}\)= (\(\sqrt{x}+\dfrac{1}{2}\))2 +\(\dfrac{3}{4}\) >\(0\) ∀ x
=> \(\dfrac{3}{x+\sqrt{x}+1}>0\) ∀ x
=> P > 0 với mọi x ≥ 0 và x ≠ 1
Cho hai bt A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)và B=\(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+2}{x-4}\)
a) rút gọn B
b)tìm x thuộc Z để C= A(B-2) có giá trị nguyên
a) \(B=\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{\sqrt{x}+2}{x-4}\left(đk:x\ge0,x\ne4\right)\)
\(=\dfrac{\sqrt{x}+\sqrt{x}+2}{x-4}.\dfrac{x-4}{\sqrt{x}+2}=\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}\)
c) \(C=A\left(B-2\right)=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\left(\dfrac{2\sqrt{x}+2}{\sqrt{x}+2}-2\right)\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}.\dfrac{-2}{\sqrt{x}+2}=\dfrac{-2}{\sqrt{x}-2}\in Z\)
\(\Rightarrow\sqrt{x}-2\inƯ\left(2\right)=\left\{1;-1;2-2\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{3;1;4;0\right\}\)
\(\Rightarrow x\in\left\{0;1;9;16\right\}\)
(P): (\(\dfrac{1}{x-\sqrt{x}}\)+\(\dfrac{1}{\sqrt{x}-1}\)):\(\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
a)rút gọn bt P
b) tìm x để P=-1
a: \(P=\dfrac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}=\dfrac{\sqrt{x}-1}{\sqrt{x}}\)
b: Để P=-1 thì \(\sqrt{x}-1=-\sqrt{x}\)
=>x=1/4(nhận)
cho bt A = \(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\) với x ≥ 0, x ≠ 4,x≠9
a, rút gọn A.
b, tính gtr của A khi x=25.
c, tìm các gtr của x để A<1.
d, tìm các gtr của x để A nhận gtr nguyên.
câu a tham khảo ở đây
https://hoc24.vn/cau-hoi/.1145652136620
b) \(x=25\Rightarrow P=\dfrac{\sqrt{25}+1}{\sqrt{25}-3}=\dfrac{6}{2}=3\)
c) \(A< 1\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}< 1\Rightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-1< 0\Rightarrow\dfrac{4}{\sqrt{x}-3}< 0\)
mà \(4>0\Rightarrow\sqrt{x}-3< 0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\Rightarrow0\le x< 9,x\ne4\)