a, Ta có : \(x=4\Rightarrow\sqrt{x}=2\)
\(\Rightarrow A=\frac{2+1}{2+2}=\frac{3}{4}\)
Vậy với x = 4 thì A = 3/4
b, \(B=\frac{3}{\sqrt{x}-1}-\frac{\sqrt{x}+5}{x-1}=\frac{3\left(\sqrt{x}+1\right)-\sqrt{x}-5}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(=\frac{3\sqrt{x}+3-\sqrt{x}-5}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2}{\sqrt{x}+1}\)( đpcm )
a, Với \(x\ge0,x\ne4\)
\(A=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}-\frac{1}{\sqrt{x}-2}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)-5-\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
b, Ta có \(x=6+4\sqrt{2}=2^2+4\sqrt{2}+\left(\sqrt{2}\right)^2=\left(2+\sqrt{2}\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(2+\sqrt{2}\right)^2}=\left|2+\sqrt{2}\right|=2+\sqrt{2}\)do \(2+\sqrt{2}>0\)
\(\Rightarrow A=\frac{2+\sqrt{2}-4}{2+\sqrt{2}-2}=\frac{-2+\sqrt{2}}{\sqrt{2}}=\frac{-2\sqrt{2}+2}{2}=\frac{-2\left(\sqrt{2}-1\right)}{2}=1-\sqrt{2}\)
1, A = \(\dfrac{\sqrt{x}-4}{\sqrt{x}-2}\)
2 , A = \(1-\sqrt{2}\)
a: Sửa đề: \(P=\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right):\dfrac{2}{x^2-2x+1}\)
\(=\left(\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}\right)\cdot\dfrac{\left(x-1\right)^2}{2}\)
\(=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)^2}\cdot\dfrac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{2}\)
\(=\dfrac{x-\sqrt{x}-2-\left(x+\sqrt{x}-2\right)}{\sqrt{x}-1}\cdot\dfrac{1}{2}\)
\(=\dfrac{-\sqrt{x}}{\sqrt{x}-1}\)
b: Để P>0 thì \(-\dfrac{\sqrt{x}}{\sqrt{x}-1}>0\)
=>\(\dfrac{\sqrt{x}}{\sqrt{x}-1}< 0\)
=>\(\sqrt{x}< 1\)
=>\(0< =x< 1\)
c: Thay \(x=7-4\sqrt{3}=\left(2-\sqrt{3}\right)^2\) vào P, ta được:
\(P=\dfrac{-\sqrt{\left(2-\sqrt{3}\right)^2}}{\sqrt{\left(2-\sqrt{3}\right)^2}-1}\)
\(=\dfrac{-\left(2-\sqrt{3}\right)}{2-\sqrt{3}-1}=\dfrac{-2+\sqrt{3}}{1-\sqrt{3}}=\dfrac{2-\sqrt{3}}{\sqrt{3}-1}\)
\(=\dfrac{\sqrt{3}-1}{2}\)
`B=(1/(3-sqrtx)-1/(3+sqrtx))*(3+sqrtx)/sqrtx(x>=0,x ne 9)`
`B=((3+sqrtx)/(9-x)-(3-sqrtx)/(9-x))*(3+sqrtx)/sqrtx`
`B=((3+sqrtx-3+sqrtx)/(9-x))*(3+sqrtx)/sqrtx`
`B=(2sqrtx)/((3-sqrtx)(3+sqrtx))*(3+sqrtx)/sqrtx`
`B=2/(3-sqrtx)`
`B>1/2`
`<=>2/(3-sqrtx)-1/2>0`
`<=>(4-3+sqrtx)/[2(3-sqrtx)]>0`
`<=>(sqrtx+1)/(2(3-sqrtx))>0`
Mà `sqrtx+1>=1>0`
`<=>2(3-sqrtx)>0`
`<=>3-sqrtx>0`
`<=>sqrtx<3`
`<=>x<9`
a: Thay x=36 vào A, ta được:
\(A=\frac{\sqrt{36}+4}{\sqrt{36}+2}=\frac{6+4}{6+2}=\frac{10}{8}=\frac54\)
b: \(B=\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right):\frac{x+16}{\sqrt{x}+2}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\cdot\frac{\sqrt{x}+2}{x+16}\)
\(=\frac{x-4\sqrt{x}+4\sqrt{x}+16}{x-16}\cdot\frac{\sqrt{x}+2}{x+16}=\frac{\sqrt{x}+2}{x-16}\)
c: Đặt P=B(A-1)
\(=\frac{\sqrt{x}+2}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\cdot\left(\frac{\sqrt{x}+4}{\sqrt{x}+2}-1\right)\)
\(=\frac{\sqrt{x}+2}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}\cdot\frac{2}{\sqrt{x}+2}=\frac{2}{x-16}\)
Để P là số nguyên thì 2⋮x-16
=>x-16∈{1;-1;2;-2}
=>x∈{17;15;18;14}
\(B=\left[\dfrac{\sqrt{x-2}}{\left(\sqrt{x}-1\right)^2}-\dfrac{\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)
\(=\left[\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)
\(=\left[\dfrac{x+\sqrt{x}-2\sqrt{x}-2-x+\sqrt{x}-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)
\(=\left[\dfrac{-2\sqrt{x}}{\left(\sqrt{x}-1\right)^2\left(\sqrt{x}+1\right)}\right]\sqrt{x}\left(\sqrt{x}-1\right)=\)
\(=\dfrac{-2x}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=-\dfrac{2x}{x-1}\)
b/
\(B=-\dfrac{2\left(x-1\right)+2}{x-1}=-2+\dfrac{2}{x-1}\)
Để B nguyên
\(x-1=\left\{-1;-2;1;2\right\}\Rightarrow x=\left[0;-1;2;3\right]\)
a: \(P=\left(\dfrac{2+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right):\dfrac{\sqrt{x}+1-\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}+1}{1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Để P nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)
\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)
hay \(x\in\left\{0;4;9\right\}\)
a, Để A nhận giá trị dương thì \(A>0\)hay \(x-1>0\Leftrightarrow x>1\)
b, \(B=2\sqrt{2^2.5}-3\sqrt{3^2.5}+4\sqrt{4^2.5}\)
\(=4\sqrt{5}-9\sqrt{5}+16\sqrt{5}=\left(4-9+16\right)\sqrt{5}=11\sqrt{5}\)
( theo công thức \(A\sqrt{B}=\sqrt{A^2B}\))
c, Với \(a\ge0;a\ne1\)
\(C=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2.\frac{1}{\left(\sqrt{a}+1\right)^2}=1\)
\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(đk:x\ge0,x\ne1\right)\)
\(=\dfrac{x+2+\sqrt{x}\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2.2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\dfrac{2}{x+\sqrt{x}+1}\)
Để A nguyên thì: \(x+\sqrt{x}+1\inƯ\left(2\right)=\left\{-2;-1;1;2\right\}\)
Mà \(x+\sqrt{x}+1=\left(x+\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{3}{4}=\left(\sqrt{x}+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
\(\Rightarrow x+\sqrt{x}+1\in\left\{1;2\right\}\)
+ Với \(x+\sqrt{x}+1=1\)
\(\Leftrightarrow\sqrt[]{x}\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow x=0\left(tm\right)\left(do.\sqrt{x}+1\ge1>0\right)\)
+ Với \(x+\sqrt{x}+1=2\)
\(\Leftrightarrow\left(x+\sqrt{x}+\dfrac{1}{4}\right)=\dfrac{5}{4}\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{2}\right)^2=\dfrac{5}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+\dfrac{1}{2}=\dfrac{\sqrt{5}}{2}\\\sqrt{x}+\dfrac{1}{2}=-\dfrac{\sqrt{5}}{2}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=\dfrac{\sqrt{5}-1}{2}\\\sqrt{x}=-\dfrac{\sqrt{5}+1}{2}\left(VLý\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{3-\sqrt{5}}{2}\left(tm\right)\)
Vậy \(S=\left\{1;\dfrac{3-\sqrt{5}}{2}\right\}\)
a,Ta có \(x=4-2\sqrt{3}=\sqrt{3}^2-2\sqrt{3}+1=\left(\sqrt{3}-1\right)^2\)
\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)do \(\sqrt{3}-1>0\)
\(\Rightarrow A=\frac{1}{\sqrt{3}-1-1}=\frac{1}{\sqrt{3}-2}\)
b, Với \(x\ge0;x\ne1\)
\(B=\left(\frac{-3\sqrt{x}}{x\sqrt{x}-1}-\frac{1}{1-\sqrt{x}}\right):\left(1-\frac{x+2}{1+\sqrt{x}+x}\right)\)
\(=\left(\frac{-3\sqrt{x}+x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{x+\sqrt{x}+1-x-2}{x+\sqrt{x}+1}\right)\)
\(=\left(\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}-1}{x+\sqrt{x}+1}\right)\)
\(=\frac{\sqrt{x}-1}{x+\sqrt{x}+1}.\frac{x+\sqrt{x}+1}{\sqrt{x}-1}=1\)
Vậy biểu thức ko phụ thuộc biến x
c, Ta có : \(\frac{2A}{B}\)hay \(\frac{2}{\sqrt{x}-1}\)để biểu thức nhận giá trị nguyên
thì \(\sqrt{x}-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
a. Ta có x=4−2√3=1−2.1.√3+3=(1−√3)2x=4−23=1−2.1.3+3=(1−3)2.
⇒√x=|1−√3|=√3−1.⇒x=|1−3|=3−1.
Khi đó A=1√x−1=1√3−1−1=1√3−2=−2−√3.A=1x−1=13−1−1=13−2=−2−3.
b. Với x≥0x≥0 và x
a) Thay x = 4 - 2\(\sqrt{3}\) vào biểu thức A = \(\dfrac{1}{\sqrt{x}-1}\)
Ta được : A = \(\dfrac{1}{\sqrt{\left(\sqrt{3}-1\right)^2}-1}\)
= \(\dfrac{1}{\sqrt{3}-1-1}\)
= \(\dfrac{1}{\sqrt{3}-2}\)
= \(\sqrt{3}+2\)
Vậy A = \(\sqrt{3}\) + 2 khi x = 4 - 2\(\sqrt{3}\)
b) B = \(\left(\dfrac{-3\sqrt{x}}{x\sqrt{x}-1}-\dfrac{1}{1-\sqrt{x}}\right)\) : \(\left(1-\dfrac{x+2}{1+\sqrt{x}+x}\right)\)
= \(\left(\dfrac{-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}-1}\right)\) : \(\left(1-\dfrac{x+2}{1+\sqrt{x}+x}\right)\)
= \(\dfrac{-3\sqrt{x}+x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) : \(\dfrac{1+\sqrt{x}+x-x-2}{1+\sqrt{x}+x}\)
= \(\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) : \(\dfrac{\sqrt{x}-1}{1+\sqrt{x}+x}\)
= \(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) . \(\dfrac{1+\sqrt{x}+x}{\sqrt{x}-1}\)
= 1 với x ≥ 0 , x ≠ 1
Vậy giá trị của B không phụ thuộc vào x
c) \(\dfrac{2A}{B}\) = \(\left(2.\dfrac{1}{\sqrt{x}-1}\right)\) : 1
= \(\dfrac{2}{\sqrt{x}-1}\) . 1
= \(\dfrac{2}{\sqrt{x}-1}\)
Để \(\dfrac{2A}{B}\) ϵ Z thì \(\dfrac{2}{\sqrt{x}-1}\)ϵ Z hay \(\sqrt{x}\) -1 ϵ Ư(2) = \(\left\{\pm1;\pm2\right\}\)
\(\sqrt{x}\) - 1 = 1 ⇔ \(\sqrt{x}\) = 2 ⇔ x = 4 (TM)
\(\sqrt{x}\) - 1 = -1 ⇔ \(\sqrt{x}\) = 0 ⇔ x = 0 (TM)
\(\sqrt{x}\) - 1 = 2 ⇔\(\sqrt{x}\) = 3 ⇔ x = 9 (TM)
\(\sqrt{x}\) - 1 = -2 ⇔ \(\sqrt{x}\) = -1 ⇔ x = \(\varnothing\) (L)
Vậy x = \(\left\{0;4;9\right\}\) thì \(\dfrac{2A}{B}\) nhận giá trị nguyên
a) Thay x = 4 - 2\(\sqrt{3}\) vào biểu thức A = \(\dfrac{1}{\sqrt{x}-1}\)
Ta được : A = \(\dfrac{1}{\sqrt{\left(\sqrt{3}-1\right)^2}-1}\)
= \(\dfrac{1}{|\sqrt{3}-1|-1}\)
= \(\dfrac{1}{\sqrt{3}-1-1}\)
= \(\dfrac{1}{\sqrt{3}-2}\)
= - 2 - \(\sqrt{3}\)
Vậy A = - 2 - \(\sqrt{3}\) khi x = 4 - 2\(\sqrt{3}\)
b) B = \(\left(\dfrac{-3\sqrt{x}}{x\sqrt{x}-1}-\dfrac{1}{1-\sqrt{x}}\right)\) : \(\left(1-\dfrac{x+2}{1+\sqrt{x}+x}\right)\)
= \(\left(\dfrac{-3\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}-1}\right)\) : \(\left(1-\dfrac{x+2}{1+\sqrt{x}+x}\right)\)
= \(\dfrac{-3\sqrt{x}+x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) : \(\dfrac{1+\sqrt{x}+x-x-2}{1+\sqrt{x}+x}\)
= \(\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) : \(\dfrac{\sqrt{x}-1}{1+\sqrt{x}+x}\)
= \(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\) . \(\dfrac{1+\sqrt{x}+x}{\sqrt{x}-1}\)
= 1 với x ≥ 0 , x ≠ 1
Vậy giá trị của B không phụ thuộc vào x
c) \(\dfrac{2A}{B}=\left(2.\dfrac{1}{\sqrt{x}-1}\right):1\)
= \(\dfrac{2}{\sqrt{x}-1}\) . 1
= \(\dfrac{2}{\sqrt{x}-1}\)
Để \(\dfrac{2A}{B}\) ϵ Z thì \(\dfrac{2}{\sqrt{x}-1}\) ϵ Z hay \(\sqrt{x}-1\) ϵ Ư(2) = \(\left\{\pm1;\pm2\right\}\)
\(\sqrt{x}-1\) = 1 ⇔ \(\sqrt{x}\) = 2 ⇔ x = 4 (TM)
\(\sqrt{x}-1\) = -1 ⇔ \(\sqrt{x}\) = 0 ⇔ x = 0 (TM)
\(\sqrt{x}-1\) = 2 ⇔ \(\sqrt{x}\) = 3 ⇔ x = 9 (TM)
\(\sqrt{x}-1\) = -2 ⇔ \(\sqrt{x}\) = -1 ⇔ x = \(\varnothing\) (L)
Vậy x = \(\left\{0;4;9\right\}\) thì \(\dfrac{2A}{B}\) nhận giá trị nguyên
a. Ta có x = 4 - 2\sqrt3 = 1 - 2.1.\sqrt3 + 3 = (1- \sqrt 3 )^2x=4−2
3
=1−2.1.
3
+3=(1−
3
)
2
.
\Rightarrow \sqrt x = |1-\sqrt 3| = \sqrt 3 - 1.⇒
x
=∣1−
3
∣=
3
−1.
Khi đó A = \dfrac1{\sqrt x - 1} = \dfrac1{\sqrt3 - 1 - 1} = \dfrac1{\sqrt3 - 2} = -2-\sqrt3.A=
x
−1
1
=
3
−1−1
1
=
3
−2
1
=−2−
3
.
b. Với x \ge 0x≥0 và x \ne 1x
=1 ta có:
B = \left(\dfrac{-3\sqrt x}{x\sqrt x - 1} - \dfrac1{1 - \sqrt x}\right):\left(1 - \dfrac{x + 2}{1 + \sqrt x + x}\right)B=(
x
x
−1
−3
x
−
1−
x
1
):(1−
1+
x
+x
x+2
)
= \left[\dfrac{-3\sqrt x}{(\sqrt x - 1)(x + \sqrt x + 1)} + \dfrac1{\sqrt x - 1}\right]:\dfrac{(1 + \sqrt x + x)-(x + 2)}{1 + \sqrt x + x}=[
(
x
−1)(x+
x
+1)
−3
x
+
x
−1
1
]:
1+
x
+x
(1+
x
+x)−(x+2)
= \dfrac{-3\sqrt x + x +\sqrt x + 1}{(\sqrt x - 1)(x + \sqrt x + 1)} . \dfrac{x + \sqrt x + 1}{\sqrt x - 1}=
(
x
−1)(x+
x
+1)
−3
x
+x+
x
+1
.
x
−1
x+
x
+1
= \dfrac{x - 2\sqrt x + 1}{(\sqrt x - 1)(x + \sqrt x + 1)} . \dfrac{x + \sqrt x + 1}{\sqrt x-1}=
(
x
−1)(x+
x
+1)
x−2
x
+1
.
x
−1
x+
x
+1
= \dfrac{(\sqrt x - 1)^2.(x + \sqrt x + 1)}{(\sqrt x - 1)^2(x + \sqrt x + 1)} = 1=
(
x
−1)
2
(x+
x
+1)
(
x
−1)
2
.(x+
x
+1)
=1
Do đó giá trị của BB không phụ thuộc vào xx.
c.Ta có \dfrac{2A}B = \dfrac{2}{\sqrt x - 1}
B
2A
=
x
−1
2
Để \dfrac{2A}B
B
2A
nhận giá trị nguyên thì \sqrt x - 1
x
−1 là ước của 22.
Suy ra \sqrt x - 1 \in \{-2;-1;1;2\}
x
−1∈{−2;−1;1;2}.
\sqrt x - 1
x
−1 -2−2 -1−1 11 22
x loại 0 4 9
Vậy x \in \{0;4;9\}x∈{0;4;9} thì \dfrac{2A}B
B
2A
nhận giá trị nguyên.
c
a. Ta có x=4−2√3=1−2.1.√3+3=(1−√3)2x=4−23=1−2.1.3+3=(1−3)2.
⇒√x=|1−√3|=√3−1.⇒x=|1−3|=3−1.
Khi đó A=1√x−1=1√3−1−1=1√3−2=−2−√3.A=1x−1=13−1−1=13−2=−2−3.
b. Với x≥0x≥0 và x≠
Đúng(0)
a. Ta có x=4−2√3=1−2.1.√3+3=(1−√3)2x=4−23=1−2.1.3+3=(1−3)2.
⇒√x=|1−√3|=√3−1.⇒x=|1−3|=3−1.
Khi đó A=1√x−1=1√3−1−1=1√3−2=−2−√3.A=1x−1=13−1−1=13−2=−2−3.
b. Với x≥0x≥0 và
a. Ta có x = 4 - 2\sqrt3 = 1 - 2.1.\sqrt3 + 3 = (1- \sqrt 3 )^2x=4−23=1−2.1.3+3=(1−3)2.
\Rightarrow \sqrt x = |1-\sqrt 3| = \sqrt 3 - 1.⇒x=∣1−3∣=3−1.
Khi đó A = \dfrac1{\sqrt x - 1} = \dfrac1{\sqrt3 - 1 - 1} = \dfrac1{\sqrt3 - 2} = -2-\sqrt3.A=x−11=3−1−11=3−21=−2−3.
b. Với x \ge 0x≥0 và x \ne 1x=1 ta có:
B = \left(\dfrac{-3\sqrt x}{x\sqrt x - 1} - \dfrac1{1 - \sqrt x}\right):\left(1 - \dfrac{x + 2}{1 + \sqrt x + x}\right)B=(
Đúng(0)
a. Ta có: x = 4 - \(2\sqrt{3}\)
<=> x = 3 - \(2\sqrt{3}\) + 1
<=> x = \(\left(\sqrt{3}-1\right)^2\)
<=> \(\sqrt{x}\) = \(\sqrt{3}-1\)
Ta có: A = \(\dfrac{1}{\sqrt{x}-1}\) với x \(\ge\) 0 và x \(\ne\) 1
Thay \(\sqrt{x}=\sqrt{3}-1\) vào A, ta có:
A = \(\dfrac{1}{\sqrt{3}-1-1}=\dfrac{1}{\sqrt{3}-2}=\dfrac{\sqrt{3}+2}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}=\dfrac{\sqrt{3}+2}{3-4}=\dfrac{\sqrt{3}+2}{-1}=-\sqrt{3}-\sqrt{2}\) (không thỏa mãn điều kiện)
Vậy khi x = 4 -\(2\sqrt{3}\) thì A không có giá trị
A = \(-2-\sqrt{3}\)
a. Ta có x = 4 - 2\sqrt3 = 1 - 2.1.\sqrt3 + 3 = (1- \sqrt 3 )^2x=4−23=1−2.1.3+3=(1−3)2.
\Rightarrow \sqrt x = |1-\sqrt 3| = \sqrt 3 - 1.⇒x=∣1−3∣=3−1.
Khi đó A = \dfrac1{\sqrt x - 1} = \dfrac1{\sqrt3 - 1 - 1} = \dfrac1{\sqrt3 - 2} = -2-\sqrt3.A=x−11=3−1−11=3−21=−2−3.
b. Với x \ge 0x≥0 và x \ne 1x=1 ta có:
B = \left(\dfrac{-3\sqrt x}{x\sqrt x - 1} - \dfrac1{1 - \sqrt x}\right):\left(1 - \dfrac{x + 2}{1 + \sqrt x + x}\right)B=(