A = ((x + 2)/(x * sqrt(x) - 1) + (sqrt(x))/(x + sqrt(x) + 1) + 1/(1 - sqrt(x))) / ((sqrt(x) - 1)/2) b) tìm x để A nguyên
Cho \(D=\left(\dfrac{x-2}{x+2}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) với \(x>0; x\ne1\)
a) Tìm x để \(2D=2\sqrt{x}+5\)
b) Tìm x để D<1
c) Tìm x nguyên để D nguyên
Cho A= \(\dfrac{\sqrt{x}+4}{{}\sqrt{x}-1}\) và B= \(\dfrac{x+2\sqrt{x}}{\sqrt{x}(\sqrt{x}+1)} -\dfrac{3\sqrt{x}-3}{x-1}\) (đk: x>0,x≠1)
a) Rút gọn P=A.B
b) Tìm x để P(\(\sqrt{x}+1\)) ≤ 6-x
c) Tìm x để P nhận giá trị nguyên
cho biểu thức
A=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) và B=\(\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
a,Tính giá trị biểu thức B khi x=36
b,Tìm x để B<\(\dfrac{1}{2}\)
c,Rút gọn A
d, Tìm giá trị x nguyên nhỏ nhất để biểu thức P=A.B nguyên
a. B = \(\dfrac{\sqrt{36}}{\sqrt{36}-3}=\dfrac{6}{6-3}=2\)
a: Thay x=36 vào B, ta được:
\(B=\dfrac{6}{6-3}=\dfrac{6}{3}=2\)
B =\(\dfrac{x-3}{x-1}-\dfrac{2}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\)
Tìm các giá trị nguyên của x để \(\dfrac{A}{B}< 1\) biết A=\(\dfrac{2\sqrt{x}-2}{\sqrt{x}+1}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >1\end{matrix}\right.\)
\(B=\dfrac{x-3}{x-1}-\dfrac{2}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{2}{\sqrt{x}+1}+\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x-3-2\left(\sqrt{x}-1\right)+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{x+\sqrt{x}-2-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(ĐặtP=\dfrac{A}{B}\)
=>\(P=\dfrac{2\sqrt{x}-2}{\sqrt{x}+1}:\dfrac{\sqrt{x}}{\sqrt{x}+1}\)
\(=\dfrac{2\sqrt{x}-2}{\sqrt{x}+1}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}}=\dfrac{2\sqrt{x}-2}{\sqrt{x}}\)
Để P<1 thì P-1<0
=>\(\dfrac{2\sqrt{x}-2-\sqrt{x}}{\sqrt{x}}< 0\)
=>\(\sqrt{x}-2< 0\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
mà x nguyên
nên \(x\in\left\{0;1;2;3\right\}\)
Cho \(A=\frac{3x+\sqrt{9x}-3}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{\sqrt{x}}\left(\frac{1}{1-\sqrt{x}}-1\right)\)
a, Rút gọn A
b, Tìm các giá trị nguyên của x để A nguyên
c, Tìm các giá trị của x để A = \(\sqrt{x}\)
cho biểu thức A=\(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\)với x≥0,x≠1
a)rút gọn A
b)tìm x nguyên để M =A.\(\dfrac{\sqrt{x}+2}{2\sqrt{x}+1}+\dfrac{x-\sqrt{x}-5}{\sqrt{x}+3}\)có giá trị nguyên
a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
4.A=\(\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right)\): \(\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)
a) Rút gọn A
b)Tìm x nguyên để A có giá trị nguyên
a. \(A=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{2\left(x-2\sqrt{x}+1\right)}{x-1}\)
\(=\left(\dfrac{\left(x\sqrt{x}-1\right)\left(x+\sqrt{x}\right)-\left(x\sqrt{x}+1\right)\left(x-\sqrt{x}\right)}{\left(x-\sqrt{x}\right)\left(x+\sqrt{x}\right)}\right):\dfrac{2\left(\sqrt{x}-1\right)^2}{x-1}\)
\(=\left(\dfrac{\left(x\sqrt{x}-1\right)\left(x+\sqrt{x}\right)-\left(x\sqrt{x}+1\right)\left(x-\sqrt{x}\right)}{x^2-x}\right).\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=\left(\dfrac{x^2\sqrt{x}+x^2-x-\sqrt{x}-\left(x^2\sqrt{x}-x^2+x-\sqrt{x}\right)}{x^2-x}\right).\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=\left(\dfrac{x^2\sqrt{x}+x^2-x-\sqrt{x}-x^2\sqrt{x}+x^2-x+\sqrt{x}}{x^2-x}\right).\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{2x^2-2x}{x^2-x}.\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=\dfrac{2\left(x^2-x\right)}{x^2-x}.\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}\)
\(=2.\dfrac{x-1}{2\left(\sqrt{x}-1\right)^2}=\dfrac{x-1}{\left(\sqrt{x}-1\right)^2}=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)^2}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b. \(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}-1+2}{\sqrt{x}-1}=1+\dfrac{2}{\sqrt{x}-1}\)
Để A có giá trị nguyên \(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\in Z\) \(\Leftrightarrow2⋮\left(\sqrt{x}-1\right)\)\(\Leftrightarrow\left(\sqrt{x}-1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)\(\Leftrightarrow\sqrt{x}\in\left\{2;0;3;-1\right\}\)
Vì \(\sqrt{x}\ge0\Rightarrow\sqrt{x}\in\left\{2;0;3\right\}\Leftrightarrow x\in\left\{4;0;9\right\}\)
Vậy để A có giá trị nguyên thì \(x\in\left\{4;0;9\right\}\)
A=\(\dfrac{3\sqrt{x}-6}{x-2\sqrt{x}}+\dfrac{\sqrt{x}-3}{\sqrt{x}}-\dfrac{1}{2-\sqrt{x}}\) và B=\(\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\)
Cho P=A.B. Tìm số nguyên x để \(\sqrt{P}< \dfrac{1}{3}\)
Ta có: \(P=A\cdot B\) (ĐK: \(x>0;x\ne4\))
\(=\left(\dfrac{3\sqrt{x}-6}{x-2\sqrt{x}}+\dfrac{\sqrt{x}-3}{\sqrt{x}}-\dfrac{1}{2-\sqrt{x}}\right)\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\right)\)
\(=\left[\dfrac{3\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right]\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\right)\)
\(=\left(\dfrac{3+\sqrt{x}-3}{\sqrt{x}}+\dfrac{1}{\sqrt{x}-2}\right)\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\right)\)
\(=\left(1+\dfrac{1}{\sqrt{x}-2}\right)\left(\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\right)\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-2}{\sqrt{x}+9}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+9}\)
Với x > 0; x ≠ 4 thì \(\sqrt{P}< \dfrac{1}{3}\Leftrightarrow P< \dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+9}< \dfrac{1}{9}\)
\(\Leftrightarrow\dfrac{\sqrt{x}-1}{\sqrt{x}+9}-\dfrac{1}{9}< 0\)
\(\Leftrightarrow\dfrac{9\left(\sqrt{x}-1\right)}{9\left(\sqrt{x}+9\right)}-\dfrac{\sqrt{x}+9}{9\left(\sqrt{x}+9\right)}< 0\)
\(\Leftrightarrow\dfrac{9\sqrt{x}-9-\sqrt{x}-9}{9\sqrt{x}+81}< 0\)
\(\Leftrightarrow\dfrac{8\sqrt{x}-18}{9\sqrt{x}+18}< 0\)
Ta thấy: \(9\sqrt{x}+18>0\forall x\)
\(\Rightarrow8\sqrt{x}-18< 0\)
\(\Rightarrow\sqrt{x}< \dfrac{18}{8}\)
\(\Rightarrow\sqrt{x}< \dfrac{9}{4}\Leftrightarrow x< \dfrac{81}{16}\)
Kết hợp với điều kiện, ta được: \(0< x\le5\)\(;x\ne4\)
\(\Rightarrow x\in\left\{1;2;3;5\right\};x\in Z\) thì \(\sqrt{P}< \dfrac{1}{3}\)
#Urushi
\(A=\frac{x-2\sqrt{x}}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}+\frac{1-2x-2\sqrt{x}}{x^2-\sqrt{x}}\)
a, Rút gọn A b, Tìm x để A nguyên
\(\text{ĐKXĐ: }x\ge0;x\ne1\)
\(A=\frac{x-2\sqrt{x}}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}+\frac{1-2x-2\sqrt{x}}{x^2-\sqrt{x}}\)
\(=\frac{x-2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}+\frac{1-2x-2\sqrt{x}}{\sqrt{x}.\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}-2x}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{1-2x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\frac{x\sqrt{x}-3x-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
ủa đề có sai ko zạy
\(=\frac{\sqrt{x}.\left(x-3\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
A=\(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)(x≥0,x≠4,x≠9)
1,Tìm x để A.\(\sqrt{x}\)=-1
2,Tìm x∈ Z để A∈Z
3, Tìm Min \(\dfrac{1}{A}\)
4,Tìm x∈N để A là số nguyên dương lớn nhất
5,Khi A+\(|A|\)=0, tìm GTLN của bth A.\(\sqrt{x}\)
1: Ta có: \(A=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}-9-\left(x-9\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{2\sqrt{x}-9-x+9+2x-4\sqrt{x}+\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
Để \(A=-\dfrac{1}{\sqrt{x}}\) thì \(x+\sqrt{x}=-\sqrt{x}+3\)
\(\Leftrightarrow x+2\sqrt{x}-3=0\)
\(\Leftrightarrow\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)=0\)
\(\Leftrightarrow x=1\left(nhận\right)\)
2: Để A nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-3\)
\(\Leftrightarrow\sqrt{x}-3\in\left\{-1;1;2;-2;4;-4\right\}\)
\(\Leftrightarrow\sqrt{x}\in\left\{2;4;5;1;7\right\}\)
\(\Leftrightarrow x\in\left\{16;25;1;49\right\}\)