Cho các biểu thức:\(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2};B=\dfrac{x-3}{x+1}\) \(\left(0\le x,x\ne9\right)\) a, Rút gọn A
b, Với P = A.B ,tìm x để P = \(\dfrac{9}{2}\)
c, Tìm x để B < 1
d, Tìm số nguyên x để P là số nguyên
\(\)A=\(\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)với B=\(\dfrac{x-3}{x+1}\)
a) rút gọn A
b) P=A.B,tìm x để P=\(\dfrac{9}{2}\)
c) tìm x để B<1
a: Ta có: \(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)
\(=\dfrac{2x^2-6x+x^2+4x+3-3+11x}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{3x^2+9x}{\left(x+3\right)\left(x-3\right)}\)
\(=\dfrac{3x}{x-3}\)
b: Ta có P=AB
nên \(P=\dfrac{3x}{x-3}\cdot\dfrac{x-3}{x+1}=\dfrac{3x}{x+1}\)
Để \(P=\dfrac{9}{2}\) thì 9x+9=6x
\(\Leftrightarrow3x=-9\)
hay x=-3(loại)
a) \(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\\ \Rightarrow A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow A=\dfrac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{3-11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow A=\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)-3+11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow A=\dfrac{2x^2-6x+x^2+4x+3-3+11x}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow A=\dfrac{3x^2+9x}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow A=\dfrac{3x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow A=\dfrac{3x}{x-3}\)
a. ĐKXĐ: \(x\ne\pm3\)
\(A=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}+\dfrac{3-11x}{9-x^2}\)
\(=\dfrac{2x}{x+3}+\dfrac{x+1}{x-3}-\dfrac{3-11x}{x^2-9}\)
\(=\dfrac{2x\left(x-3\right)+\left(x+1\right)\left(x+3\right)-\left(3-11x\right)}{x^2-9}\)
\(=\dfrac{2x^2-6x+x^2+4x+3-3+11x}{x^2-9}\)
\(=\dfrac{3x^2+9x}{x^2-9}=\dfrac{3x\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x}{x-3}\)
b. \(P=A.B\)
\(\Rightarrow P=\dfrac{3x}{x-3}.\dfrac{x-3}{x+1}=\dfrac{3x}{x+1}\)
Ta có \(P=\dfrac{9}{2}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\\dfrac{3x}{x+1}=\dfrac{9}{2}\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\6x=9x+9\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\-3x=9\end{matrix}\right.\) \(\Leftrightarrow x=-3\)
c. \(B< 1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\\dfrac{x-3}{x-1}< 1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\\dfrac{x-3}{x-1}-1< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\\dfrac{2}{1-x}< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\1-x< 0\end{matrix}\right.\) \(\Leftrightarrow x>1\)
Cho biểu thức D = \(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
với \(x\ne9,x\ge0\)
a) Rút gọn D
b)Tìm x để \(D< \dfrac{-1}{4}\)
a) \(D=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
\(=\dfrac{-3\sqrt{x}+3}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}-1}=\dfrac{-3}{\sqrt{x}+3}\)
b) \(D=-\dfrac{3}{\sqrt{x}+3}< -\dfrac{1}{4}\)
\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow\sqrt{x}< 9\)
\(\Leftrightarrow0\le x< 81\) và \(x\ne9\)
a) D=\(\left(\dfrac{2\sqrt{x}.\left(\sqrt{x}-3\right)+\sqrt{x}.\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-3\right)}\right)\) \(:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(\Leftrightarrow D=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}\) \(.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3-3\sqrt{x}}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3.\left(\sqrt{x}+1\right)}{\sqrt{x}+3}.\dfrac{1}{\sqrt{x}+1}\)
\(\Leftrightarrow D=\dfrac{-3}{\sqrt{x}+3}\)
b) Để D\(< \dfrac{-1}{4}\) \(\Leftrightarrow\dfrac{-3}{\sqrt{x}+3}< \dfrac{-1}{4}\)
\(\Leftrightarrow12>\sqrt{x}+3\Leftrightarrow9>\sqrt{x}\Leftrightarrow81>x\ge0\)
1) cho biểu thức A= \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{9-x}\)
B= \(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\) với \(0\le x\ne9\)
b) rút gọn A
c) tìm số nguyên x để P= A. B là số nguyên
giúp mk vs ah mk cần gấp lắm
\(A=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)\(A=\dfrac{2x-6\sqrt{x}+x+\sqrt{x+}3\sqrt{x}+3+3-11\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)\(A=\dfrac{3x-13\sqrt{x}+6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
Cho biểu thức:\(M=\left(\dfrac{3}{\sqrt{x}+3}+\dfrac{x+9}{x-9}\right):\left(\dfrac{2\sqrt{x}-5}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\) với: \(x>0;x\ne9\)
1/ Rút gọn biểu thức M |
2/ Tìm x sao cho M < 0 |
3/ Tìm số tự nhiên x để M nguyên âm |
4/ Cho x > 4. Tìm giá trị nhỏ nhất của M |
a) \(M=\left(\dfrac{3}{\sqrt{x}+3}+\dfrac{x+9}{x-9}\right):\left(\dfrac{2\sqrt{x}-5}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
\(=\dfrac{3.\left(\sqrt{x}-3\right)+x+9}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-5-\left(\sqrt{x}-3\right)}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}:\dfrac{\sqrt{x}-2}{\sqrt{x}.\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right).\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}.\left(\sqrt{x}-3\right)}{\sqrt{x}-2}=\dfrac{x}{\sqrt{x}-2}\)
b) \(M< 0\Leftrightarrow\sqrt{x}-2< 0\Leftrightarrow x< 4\)
Kết hợp điều kiện ta được \(0< x< 4\) thì M < 0
c) Từ câu b ta có M < 0 \(\Leftrightarrow0< x< 4\)
nên \(x\inℤ\) để M nguyên âm <=> \(x\in\left\{1;2;3\right\}\)
Thay lần lượt các giá trị vào M được x = 1 thỏa
d) \(M=\dfrac{x}{\sqrt{x}-2}=\sqrt{x}+2+\dfrac{4}{\sqrt{x}-2}=\left(\sqrt{x}-2+\dfrac{4}{\sqrt{x}-2}\right)+4\)
Vì x > 4 nên \(\sqrt{x}-2>0\)
Áp dụng BĐT Cauchy ta có
\(M=\left(\sqrt{x}-2+\dfrac{4}{\sqrt{x}-2}\right)+4\ge2\sqrt{\left(\sqrt{x}-2\right).\dfrac{4}{\sqrt{x}-2}}+4=8\)
Dấu "=" xảy ra khi \(\sqrt{x}-2=\dfrac{4}{\sqrt{x}-2}\Leftrightarrow x=16\left(tm\right)\)
1) \(M=\left(\dfrac{3}{\sqrt[]{x}+3}+\dfrac{x+9}{x-9}\right):\left(\dfrac{2\sqrt[]{x}-5}{x-3\sqrt[]{x}}-\dfrac{1}{\sqrt[]{x}}\right)\left(x>0;x\ne9\right)\)
\(\Leftrightarrow M=\left(\dfrac{3\left(\sqrt[]{x}-3\right)}{\left(\sqrt[]{x}+3\right)\left(\sqrt[]{x}-3\right)}+\dfrac{x+9}{x-9}\right):\left(\dfrac{2\sqrt[]{x}-5}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}-\dfrac{1}{\sqrt[]{x}}\right)\)
\(\Leftrightarrow M=\left(\dfrac{3\sqrt[]{x}-9+x+9}{x-9}\right):\left(\dfrac{2\sqrt[]{x}-5-\left(\sqrt[]{x}-3\right)}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}\right)\)
\(\Leftrightarrow M=\left(\dfrac{3\sqrt[]{x}+x}{x-9}\right):\left(\dfrac{2\sqrt[]{x}-5-\sqrt[]{x}+3}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}\right)\)
\(\Leftrightarrow M=\left(\dfrac{\sqrt[]{x}\left(\sqrt[]{x}+3\right)}{x-9}\right):\left(\dfrac{\sqrt[]{x}-2}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}\right)\)
\(\Leftrightarrow M=\left(\dfrac{\sqrt[]{x}}{\sqrt[]{x}-3}\right):\left(\dfrac{\sqrt[]{x}-2}{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}\right)\)
\(\Leftrightarrow M=\dfrac{\sqrt[]{x}}{\sqrt[]{x}-3}.\dfrac{\sqrt[]{x}\left(\sqrt[]{x}-3\right)}{\sqrt[]{x}-2}\)
\(\Leftrightarrow M=\dfrac{x}{\sqrt[]{x}-2}\)
2) Để \(M< 0\) khi và chỉ chi
\(M=\dfrac{x}{\sqrt[]{x}-2}< 0\left(1\right)\)
Nghiệm của tử là \(x=0\)
Nghiệm của mẫu \(\sqrt[]{x}-2=0\Leftrightarrow\sqrt[]{x}=2\Leftrightarrow x=4\)
Lập bảng xét dấu... ta được
\(\left(1\right)\Leftrightarrow0< x< 4\)
3) \(M=\dfrac{x}{\sqrt[]{x}-2}\inℤ^-\)
\(\Leftrightarrow x⋮\sqrt[]{x}-2\)
\(\Leftrightarrow x-\sqrt[]{x}\left(\sqrt[]{x}-2\right)⋮\sqrt[]{x}-2\)
\(\Leftrightarrow x-x+2\sqrt[]{x}⋮\sqrt[]{x}-2\)
\(\Leftrightarrow2\sqrt[]{x}⋮\sqrt[]{x}-2\)
\(\Leftrightarrow2\sqrt[]{x}-2\left(\sqrt[]{x}-2\right)⋮\sqrt[]{x}-2\)
\(\Leftrightarrow2\sqrt[]{x}-2\sqrt[]{x}+4⋮\sqrt[]{x}-2\)
\(\Leftrightarrow4⋮\sqrt[]{x}-2\)
\(\Leftrightarrow\sqrt[]{x}-2\in\left\{-1;-2;-4\right\}\)
\(\Leftrightarrow x\in\left\{1;0\right\}\)
1/ Cho hai biểu thức A=\(\dfrac{\sqrt{x}-2}{\sqrt{x}}\) và B=\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}\)+ \(\dfrac{\sqrt{x}}{\sqrt{x}-3}\)-\(\dfrac{2x}{x-9}\) với x>0 , x≠9
a) Rút gọn biểu thức B
b) Tìm các giá trị nguyên của x để P<0 với P=A.B
Lời giải:
a.
\(B=\frac{2\sqrt{x}(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}+3)-2x}{(\sqrt{x}+3)(\sqrt{x}-3)}=\frac{x-3\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}=\frac{\sqrt{x}(\sqrt{x}-3)}{(\sqrt{x}+3)(\sqrt{x}-3)}=\frac{\sqrt{x}}{\sqrt{x}+3}\)
b.
\(P=AB=\frac{\sqrt{x}-2}{\sqrt{x}}.\frac{\sqrt{x}}{\sqrt{x}+3}=\frac{\sqrt{x}-2}{\sqrt{x}+3}\)
Để $P<0\Leftrightarrow \frac{\sqrt{x}-2}{\sqrt{x}+3}<0$
Mà $\sqrt{x}+3>0$ nên $\sqrt{x}-2<0$
$\Leftrightarrow 0< x< 4$
Kết hợp với ĐKXĐ suy ra $0< x< 4$
Mà $x$ nguyên nên $x\in left\{1; 2; 3\right\}$
A=\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3\left(\sqrt{x}+3\right)}{x-9}\right)\)\(:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)(với \(x\ge0;x\ne9\))
a) Rút gọn A
b) Tìm x để A<\(-\)1
\(a,A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{x-6\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
Cho biểu thứ :\(P:\left(\dfrac{x-1}{x-3}+\dfrac{2}{x-3}+\dfrac{x^2+3}{9-x^2}\right):\left(\dfrac{2x-1}{2x+1-1}\right)\)
a) Rút gọn biểu thức P
b) Tính giá trị của P biết \(\left|x+1\right|=\dfrac{1}{2}\)
c) Tìm x để \(P=\dfrac{x}{2}\)
d) Tìm giá trị nguyen của x để P có giá trị nguyên
Cho các biểu thức sau (giải chi tiết)
A = \(\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}\) và B = \(\dfrac{2x+3\sqrt{x}+9}{x-9}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\) với \(x\ge0;x\ne9\)
a) Rút gọn biểu thức B
b) Cho \(P=\dfrac{A}{B}\). Tìm GTNN của P
a: \(B=\dfrac{2x+3\sqrt{x}+9-x+3\sqrt{x}}{x-9}=\dfrac{x+9}{x-9}\)
b: \P=A:B
\(=\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}\cdot\dfrac{x-9}{x+9}=\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}{x+9}>=\dfrac{-1\cdot3}{9}=\dfrac{-1}{3}\)
Dấu = xảy ra khi x=0
Cho biểu thức:
\(B=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}+\dfrac{2}{\sqrt{x}}\right)\)
với x > 0 , x ≠ 9
a. Rút gọn B
b. Tìm x để B < 0