Bài 8:Cho A=\(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)và B=\(\dfrac{\sqrt{x}+3}{\sqrt{x}+1}-\dfrac{5}{1-\sqrt{x}}+\dfrac{4}{x-1}\)(x≥0;x≠1)
a)Tính giá trị của A khi x=\(4+2\sqrt{3}\)
b)Rút gọn B
c)Tìm x để P=A.B có giá trị nguyên
Bài 1: Cho biểu thức A = 1 - \(\dfrac{\sqrt{x}}{1+\sqrt{x}}\), B = \(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\)+ \(\dfrac{\sqrt{x}+2}{3-\sqrt{x}}\)- \(\dfrac{10-5\sqrt{x}}{x-5\sqrt{x}+6}\)
(với x ≥ 0, x ≠ 4, x ≠ 9)
a, Tính giá trị của A biết x = 6-2\(\sqrt{5}\)
b, Rút gọn P = A : B
c, Tìm giá trị nhỏ nhất của P
a: Thay \(x=6-2\sqrt{5}\) vào A, ta được:
\(A=1-\dfrac{\sqrt{5}-1}{\sqrt{5}-1+1}=1-\dfrac{\sqrt{5}-1}{\sqrt{5}}=\dfrac{\sqrt{5}}{5}\)
b: Ta có: P=A:B
\(=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{5\sqrt{x}-10}{x-5\sqrt{x}+6}\right)\)
\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{x-4\sqrt{x}+3-x+4+5\sqrt{x}-10}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{1}{\sqrt{x}+1}:\dfrac{1}{\sqrt{x}-2}\)
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
1)so sánh 2 số sau M=\(\sqrt{18}-\sqrt{8}\) và N=\(\dfrac{5+\sqrt{5}}{\sqrt{5}+1}-\sqrt{6-2\sqrt{5}}\)
2)cho biểu thức A=\((\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}):(\dfrac{x-4}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}})\) với x>0,\(x\ne4\),\(x\ne9\)
câu 2 rút gọn A và tìm các giá trị nguyên của x để A nhận giá trị âm
1) So sánh:
N = \(\dfrac{5+\sqrt{5}}{\sqrt{5}+1}-\sqrt{6-2\sqrt{5}}\)
\(=\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}-\sqrt{\left(\sqrt{5}-1\right)^2}\)
\(=\sqrt{5}-\left(\sqrt{5}-1\right)=1\)
M = \(\sqrt{18}-\sqrt{8}\)
\(=3\sqrt{2}-2\sqrt{2}\)
\(=\sqrt{2}\)
Ta có: \(1=\sqrt{1}\)
Mà 1 < 2
\(\Rightarrow\sqrt{1}< \sqrt{2}\)
Hay 1 \(< \sqrt{2}\)
Vậy N < M
2) Với \(x>0;x\ne4;x\ne9\), ta có:
A = \(\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{x-4}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\)
\(=\left[\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\left[\dfrac{x-4}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\right]\)
\(=\dfrac{x-3\sqrt{x}-2x}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{x-4-2\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x-3}\right)}\)
\(=\dfrac{-x-3\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-2\sqrt{x}+2}\)
\(=\dfrac{-\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{x-2\sqrt{x}+2}\)
\(=\dfrac{-x}{x-2\sqrt{x}+2}\)
Bài 4: Tính và rút gọn
C = \(\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}\)
D = \(\dfrac{1}{x-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}+\dfrac{1}{x+\sqrt{x}}\:\:\left(x\ne1;x>0\right)\)
\(C=\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}\)
\(=\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{9-5}}\)
\(=\dfrac{2}{\sqrt{5}+1}+\sqrt{\dfrac{6+2\sqrt{5}}{4}}\)
\(=\dfrac{2\left(\sqrt{5}-1\right)}{5-1}+\dfrac{\sqrt{6+2\sqrt{5}}}{2}\)
\(=\dfrac{\sqrt{5}-1}{2}+\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{2}\)
\(=\dfrac{\sqrt{5}-1}{2}+\dfrac{\sqrt{5}+1}{2}=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{2}=\dfrac{2\sqrt{5}}{2}=\sqrt{5}\)
\(D=\dfrac{1}{x-\sqrt{x}}-\dfrac{2\sqrt{x}}{x-1}+\dfrac{1}{x+\sqrt{x}}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+1-2x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)\cdot\sqrt{x}}\)
\(=\dfrac{-2x+2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{-2\left(x-\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(\sqrt{x}+1\right)}=-\dfrac{2}{\sqrt{x}+1}\)
Bài 1:Tìm ĐKXĐ:
a.\(\sqrt{3x}\)
b.\(\sqrt{\dfrac{x-1}{x+3}}\)
Bài 2:Thực hiện phép tính:
C=\(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
Bài 3:
A=(1-\(\dfrac{\sqrt{x}}{\sqrt{x}+2}\)):(\(\dfrac{1}{\sqrt{x}-2}-\dfrac{2}{x-4}\)) với x>0;x≠4
a.Rút gọn A
b.Tính giá trị của A khi x =\(\dfrac{1}{4}\)
c. Chứng minh A<2
d.Tìm giá trị nguyên của x để A nguyên.
Trả lời giúp mình với ạ!Mình cảm ơn nhiều!
Bài 1:
a. ĐKXĐ: $3x\geq 0$
$\Leftrightarrow x\geq 0$
b. ĐKXĐ: $\frac{x-1}{x+3}\geq 0$
\(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x-1\geq 0\\ x+3>0\end{matrix}\right.\\ \left\{\begin{matrix} x-1\leq 0\\ x+3< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x\geq 1\\ x< -3\end{matrix}\right.\)
Bài 2:
\(C=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{2+2\sqrt{2.3}+3}-\sqrt{2-2\sqrt{2.3}+3}\)
\(=\sqrt{(\sqrt{2}+\sqrt{3})^2}-\sqrt{(\sqrt{2}-\sqrt{3})^2}\)
\(=|\sqrt{2}+\sqrt{3}|-|\sqrt{2}-\sqrt{3}|=(\sqrt{2}+\sqrt{3})-(\sqrt{3}-\sqrt{2})\)
\(=2\sqrt{2}\)
Bài 3:
a.
\(A=\frac{2}{\sqrt{x}+2}:\left[\frac{\sqrt{x}+2}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{2}{(\sqrt{x}-2)(\sqrt{x}+2)}\right]\)
\(=\frac{2}{\sqrt{x}+2}:\frac{\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{2}{\sqrt{x}+2}.\frac{(\sqrt{x}-2)(\sqrt{x}+2)}{\sqrt{x}}=\frac{2(\sqrt{x}-2)}{\sqrt{x}}\)
b. Khi $x=\frac{1}{4}$ thì $\sqrt{x}=\frac{1}{2}$.
Khi đó $A=\frac{2(\frac{1}{2}-2)}{\frac{1}{2}}=-6$
c.
$A=\frac{2(\sqrt{x}-2)}{\sqrt{x}}=2-\frac{4}{\sqrt{x}}$
$< 2$ do $\frac{4}{\sqrt{x}}>0$
Ta có đpcm
d. Với $x$ nguyên, để $A$ nguyên thì $\sqrt{x}$ là ước của $4$
$\Leftrightarrow \sqrt{x}\in\left\{1;2;4\right\}$
$\Rightarrow x\in\left\{1;4;16\right\}$ (đều tm)
Bài : Tính
a)A= \(\dfrac{2}{\sqrt{2}-1}\) - \(\sqrt{\dfrac{3}{4}}\) : \(\sqrt{\dfrac{3}{2}}\) +\(\sqrt{8}\)
b)B= \(\sqrt{9x^2}\) +\(\sqrt{x}\) (\(\sqrt{x}\) +1) với x > 0
a: \(=\dfrac{2\left(\sqrt{2}+1\right)}{2-1}-\sqrt{\dfrac{3}{4}:\dfrac{3}{2}}+2\sqrt{2}\)
\(=2\sqrt{2}+2+2\sqrt{2}-\sqrt{\dfrac{1}{2}}\)
\(=4\sqrt{2}+2-\dfrac{\sqrt{2}}{2}=\dfrac{7}{2}\sqrt{2}+2\)
b: \(B=\left|3x\right|+x+\sqrt{x}\)
\(=3x+x+\sqrt{x}=4x+\sqrt{x}\)
Bài 1 :Cho hai biểu thức\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) và\(B=\dfrac{3}{\sqrt{x}-1}-\dfrac{\sqrt{x}+5}{x-1}\) với x≥ 0; x≠1
a. Tính giá trị của biểu thức A khi x = 4
b. Chứng minh\(\dfrac{2}{\sqrt{x}+1}\)
Bài 2:
Cho biểu thức:\(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
Rút gọn P
Bài 2:
Ta có: \(P=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+1}{\sqrt{x}+3}\)
Bài 3: Tìm x biết:
a) \(\sqrt{3x-2}=4\)
b) \(\sqrt{4x^2+4x+1}-11=5\)
Bài 4: Cho biểu thức
C= \(1\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\) (x > 0, x ≠ 1)
a) Rút gọn C
b) Tìm x để C - 6 < 0
Helpp!!!
Bài 3:
a) \(\sqrt{3x-2}=4\)
⇔\(\sqrt{3x-2}=\sqrt{4^2}\)
⇔\(3x-2=4^2=16\)
\(3x=16+2=18\)
\(x=18:3=6\)
Vậy \(x=6\)
b)\(\sqrt{4x^2+4x+1}-11=5\)
⇔\(\sqrt{\left(2x\right)^2+2\left(2x\right)\cdot1+1^2}-11=5\)
⇔\(\sqrt{\left(2x+1\right)^2}-11=5\)
TH1:
⇔\(\left(2x+1\right)-11=5\)
\(2x+1=5+11=16\)
\(2x=16-1=15\)
\(x=15:2=7,5\)
TH2:
⇔\(\left(2x+1\right)-11=-5\)
\(2x-1=-5+11=6\)
\(2x=6+1=7\)
\(x=7:2=3,5\)
Vậy \(x=\left\{7,5;3,5\right\}\)
(Câu này mình không chắc chắn lắm)
(Học sinh lớp 6 đang làm bài này)
Bài 4:
a: \(C=\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)
\(=\dfrac{x-1}{\sqrt{x}}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)+\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-\sqrt{x}+x+\sqrt{x}}{\sqrt{x}}=\dfrac{2x}{\sqrt{x}}=2\sqrt{x}\)
b: C-6<0
=>C<6
=>\(2\sqrt{x}< 6\)
=>\(\sqrt{x}< 3\)
=>0<=x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0< x< 9\\x\ne1\end{matrix}\right.\)
Bài 3
a)\(\sqrt{3x-2}=4\Leftrightarrow3x-2=16\Leftrightarrow3x=18\Leftrightarrow x=6\)
Vậy PT có nghiệm x=6
b)\(\sqrt{4x^2+4x+1}-11=5\Leftrightarrow\sqrt{\left(2x+1\right)^2}=16\Leftrightarrow2x+1=16hoặc2x+1=-16\)
+)TH1: \(2x+1=16\Leftrightarrow x=\dfrac{15}{2}\Leftrightarrow x=7,5\)
+)TH2:\(2x+1=-16\Leftrightarrow x=\dfrac{17}{2}\Leftrightarrow x=8,5\)
Bài 4
a)\(C=1\left(\sqrt{x}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}+\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\Leftrightarrow C=\dfrac{x-1}{\sqrt{x}}\left(\dfrac{x-\sqrt{x}+x+\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\Leftrightarrow C=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}}\dfrac{2x}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\Leftrightarrow C=\dfrac{2x}{\sqrt{x}}\Leftrightarrow C=2\sqrt{x}\)
\(Vậy\) \(C=2\sqrt{x}\)
Bài 1: cho biểu thức
P = \(\dfrac{x+3}{\sqrt{x}-2}\) và Q = \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}\) với x>0, x≠4
Ta có : \(P=\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)ĐK : \(x\ge0;x\ne4\)
Thay x = 9 vào P ta được
\(P=\dfrac{\sqrt{9}+3}{\sqrt{9}-2}=\dfrac{6}{1}=6\)
Với \(x>0;x\ne4\)
\(Q=\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)+5\sqrt{x}-2}{x-4}\)
\(=\dfrac{x-3\sqrt{x}+2+5\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{x-4}=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
a : \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)với a ≥ 0 x ≠ 4
b : \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}}\)
c : \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
d : \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\)
a) \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\) \(\left(x\ge0;x\ne4\right)\)
\(=\dfrac{x}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+2\sqrt{x}}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}}{\sqrt{x}-2}\)
b) \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\cdot\dfrac{\sqrt{x}}{x+\sqrt{x}}\) (\(x>0\))
\(=\left[\dfrac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{x}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\cdot\dfrac{\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)^2}\)
\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(x+2\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+1}{x\sqrt{x}+2x+\sqrt{x}}\)
c) \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\) (\(x\ge0;x\ne1\))
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}+\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}+3\sqrt{x}-3-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
d) \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\) \(\left(a\ne1;a\ge0\right)\)
\(=\left[\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\dfrac{\sqrt{a}+1+\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\left(\sqrt{a}+1\right)^2-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}:\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{a+2\sqrt{a}+1-a\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{2\sqrt{a}}\)
\(=\dfrac{a-a\sqrt{a}+2\sqrt{a}+1}{2\sqrt{a}}\)