A=\(\sqrt{3-x}\)+x
Rút gọn biểu thức:
a, \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
b, \(\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
a: Ta có: \(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
b: Ta có: \(\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\sqrt{a}\left(\sqrt{a}+1\right)-\left(2\sqrt{a}-1\right)+1\)
\(=a+\sqrt{a}-2\sqrt{a}+1+1\)
\(=a-\sqrt{a}+2\)
a,ĐKXĐ: tự tìm :v
\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(x+2\sqrt{x}+1\right)-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}+1\right)^2-4}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}+\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}+3\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\dfrac{3x+7\sqrt{x}-6}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}+\dfrac{2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6+2x+\sqrt{x}-3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{9\sqrt{x}-x-8}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(9\sqrt{x}-9\right)-\left(x-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{9\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(10-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(\dfrac{10-\sqrt{x}}{\sqrt{x}+3}\)
A =\(\dfrac{x\sqrt[]{x}-3}{x-2\sqrt[]{x}-3}-\dfrac{2\left(\sqrt[]{x}-3\right)}{\sqrt[]{x}+1}+\dfrac{\sqrt[]{x}+3}{3-\sqrt[]{x}}\)
a. rút gọn A
b. Tính A với x = \(14-6\sqrt[]{5}\)
c. tìm min A
a: Ta có: \(A=\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\)
\(=\dfrac{x\sqrt{x}-3-2\left(x-6\sqrt{x}+9\right)-x-4\sqrt{x}-3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x\sqrt{x}-x-4\sqrt{x}-6-2x+12\sqrt{x}-18}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x\sqrt{x}-3x+8\sqrt{x}-24}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x\left(\sqrt{x}-3\right)+8\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x+8}{\sqrt{x}+1}\)
A=\(\dfrac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}+3}{3-\sqrt{x}}\)
a) Rút gọn A
b) Tính A với x=14-6\(\sqrt{5}\)
c) Tìm Min A
Lời giải:
ĐKXĐ: $x\geq 0; x\neq 9$
a. \(A=\frac{x\sqrt{x}-3}{(\sqrt{x}+1)(\sqrt{x}-3)}-\frac{2(\sqrt{x}-3)^2}{(\sqrt{x}+1)(\sqrt{x}-3)}-\frac{(\sqrt{x}+3)(\sqrt{x}+1)}{(\sqrt{x}+1)(\sqrt{x}-3)}\)
\(=\frac{x\sqrt{x}-3x+8\sqrt{x}-24}{(\sqrt{x}+1)(\sqrt{x}-3)}=\frac{(\sqrt{x}-3)(x+8)}{(\sqrt{x}+1)(\sqrt{x}-3)}=\frac{x+8}{\sqrt{x}+1}\)
b.
\(14-6\sqrt{5}=(3-\sqrt{5})^2\Rightarrow \sqrt{x}=3-\sqrt{5}\)
\(A=\frac{14-6\sqrt{5}+8}{3-\sqrt{5}+1}=\frac{22-6\sqrt{5}}{4-\sqrt{5}}=\frac{58-2\sqrt{5}}{11}\)
c.
Áp dụng BĐT Cô-si:
$x+4\geq 4\sqrt{x}\Rightarrow x+8\geq 4(\sqrt{x}+1)$
$\Rightarrow A=\frac{x+8}{\sqrt{x}+1}\geq 4$
Vậy $A_{\min}=4$. Giá trị này đạt tại $x=4$
28. A=\(\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\)
a. rút gọn A
b. tính A với x = \(7-4\sqrt{3}\)
c. tìm x khi A=3
a:
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-1\right):\dfrac{9-x+x-9-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{\sqrt{x}-\sqrt{x}-3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{-\left(\sqrt{x}-2\right)^2}=\dfrac{3}{\sqrt{x}-2}\)
b: Khi x=7-4căn 3 thì
\(A=\dfrac{3}{2-\sqrt{3}-2}=\dfrac{3}{-\sqrt{3}}=-\sqrt{3}\)
c: A=3
=>căn x-2=1
=>x=9(loại)
\(a,A=\left(\dfrac{x-3\sqrt{x}}{x-9}-1\right):\left(\dfrac{9-x}{x+\sqrt{x}-6}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\right)\left(dkxd:x\ne4,x\ge0,x\ne9\right)\)
\(=\dfrac{x-3\sqrt{x}-x+9}{x-9}:\dfrac{9-x+\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-3\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{9-x+x-9-x+4\sqrt{x}-4}\)
\(=\dfrac{-3\left(\sqrt{x}-3\right)}{\sqrt{x}-3}.\dfrac{\sqrt{x}-2}{4\sqrt{x}-4-x}\)
\(=\dfrac{-3\left(\sqrt{x}-2\right)}{-\left(x-4\sqrt{x}+4\right)}\)
\(=\dfrac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)^2}\)
\(=\dfrac{3}{\sqrt{x}-2}\)
\(b,x=7-4\sqrt{3}\Rightarrow A=\dfrac{3}{\sqrt{7-4\sqrt{3}}-2}=\dfrac{3}{\sqrt{\left(\sqrt{3}-2\right)^2}-2}=\dfrac{3}{\left|\sqrt{3}-2\right|-2}=\dfrac{3}{-\sqrt{3}+2-2}=\dfrac{\sqrt{3^2}}{-\sqrt{3}}=-\sqrt{3}\)
\(c,A=3\Rightarrow\dfrac{3}{\sqrt{x}-2}=3\\ \Rightarrow\dfrac{3-3\left(\sqrt{x}-2\right)}{\sqrt{x}-2}=0\\ \Rightarrow3-3\sqrt{x}+6=0\\ \Rightarrow-3\sqrt{x}=-9\\ \Rightarrow\sqrt{x}=3\\ \Rightarrow x=9\left(ktm\right)\)
Vậy không có giá trị x thỏa mãn đề bài.
\(\left(\frac{\sqrt{x}}{1-\sqrt{x}}+\frac{\sqrt{x}}{1+\sqrt{x}}\right)+\frac{3-\sqrt{x}}{x-1}\)
\(\left(\frac{3+\sqrt{x}}{3-\sqrt{x}}-\frac{3-\sqrt{x}}{3+\sqrt{x}}-\frac{4x}{x-9}\right):\left(\frac{5}{3-\sqrt{x}}-\frac{4\sqrt{x+2}}{3\sqrt{x}-x}\right)\)
\(\left(1+\frac{\sqrt{a}}{a+1}\right):\left(\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}-a-1}\right)\)
\(\frac{3a-3+\sqrt{9a}}{a+\sqrt{a}-2}-\frac{\sqrt{a}+1}{\sqrt{a}+2}+\frac{\sqrt{a-2}}{1-\sqrt{a}}\)
câu cuối sai nhé . đúng thì ntn
\(\frac{3a-3+\sqrt{9a}}{a+\sqrt{a-2}}-\frac{\sqrt{a+1}}{\sqrt{a+2}}+\frac{\sqrt{a}-2}{1-\sqrt{a}}\)
Cô giúp em nhé :)
a. \(A=\frac{\sqrt{x}\left(\sqrt{x}+1\right)+\sqrt{x}\left(\sqrt{x}-1\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\frac{3-\sqrt{x}}{x-1}\)
\(=\frac{x+\sqrt{x}+x-\sqrt{x}}{1-x}+\frac{3-\sqrt{x}}{x-1}=\frac{-2x-\sqrt{x}+3}{x-1}\)
\(=\frac{\left(-2\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{-2\sqrt{x}-3}{\sqrt{x}+1}\)
b. \(B=\frac{\left(3+\sqrt{x}\right)^2-\left(3-\sqrt{x}\right)^2+4x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\frac{5\sqrt{x}-4\sqrt{x}-2}{\sqrt{x}\left(3-\sqrt{x}\right)}\)
\(B=\frac{12\sqrt{x}+4x}{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}:\frac{\sqrt{x}-2}{\sqrt{x}\left(3-\sqrt{x}\right)}=\frac{4x}{\sqrt{x}-2}\)
Cho biểu thức: \(A\) = \(\left(\dfrac{3+\sqrt{x}}{3-\sqrt{x}}-\dfrac{3-\sqrt{x}}{3+\sqrt{x}}-\dfrac{4x}{x-9}\right)\) : \(\left(\dfrac{5}{3-\sqrt{x}}-\dfrac{4\sqrt{x}+2}{3\sqrt{x}-x}\right)\) . Tìm đk của x để |A| > - A
ĐKXĐ: x>0; x<>9
\(A=\left(\dfrac{-\left(\sqrt{x}+3\right)}{\sqrt{x}-3}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}-\dfrac{4x}{x-9}\right):\left(\dfrac{5\sqrt{x}-4\sqrt{x}-2}{\sqrt{x}\left(3-\sqrt{x}\right)}\right)\)
\(=\dfrac{-x-6\sqrt{x}-9+x-6\sqrt{x}+9-4x}{x-9}:\dfrac{-\sqrt{x}+2}{\sqrt{x}\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-4x-12\sqrt{x}}{x-9}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{-\left(\sqrt{x}-2\right)}\)
\(=\dfrac{4x\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}{\left(x-9\right)\left(\sqrt{x}-2\right)}=\dfrac{4x}{\sqrt{x}-2}\)
|A|>-A
=>A>=0
=>4x>0
=>x>0 và x<>9
Cho \(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
a) Tìm x để A=1
b) Tính A với \(x=4-2\sqrt{3}\)
c) Tìm x để 5A nguyên
a: Ta có: \(A=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\dfrac{3\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
\(=\dfrac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{-5\sqrt{x}+2}{\sqrt{x}+3}\)
Phân tích thành nhân tử
\(x+\sqrt{x}\)
\(x-\sqrt{x}\)
\(a+3\sqrt{a}-10\)
\(x\sqrt{x}+\sqrt{x}-x-1\)
\(x+\sqrt{x}-2\)
\(x-5\sqrt{x}+6\)
\(x\sqrt{x}-1\)
\(x\sqrt{x}-x+\sqrt{x}-1\)
\(x+2\sqrt{x}-15\)
\(x-2\sqrt{x}-3\)
\(a+\sqrt{a}-6\)
\(x-16\)
\(x+2\sqrt{x}+1\)
\(x-1\)
\(x-2\sqrt{x}+1\)
\(a\sqrt{a}+1\)
\(a+\sqrt{a}-2\)
\(2x-5\sqrt{x}+3\)
\(x-9\)
\(x+\sqrt{x}-6\)
1. $x+\sqrt{x}=\sqrt{x}(\sqrt{x}+1)$
2. $x-\sqrt{x}=\sqrt{x}(\sqrt{x}-1)$
3. $a+3\sqrt{a}-10=(a-2\sqrt{a})+(5\sqrt{a}-10)$
$=\sqrt{a}(\sqrt{a}-2)+5(\sqrt{a}-2)=(\sqrt{a}+5)(\sqrt{a}-2)$
4. $x\sqrt{x}+\sqrt{x}-x-1=(x\sqrt{x}+\sqrt{x})-(x+1)=\sqrt{x}(x+1)-(x+1)$
$=(x+1)(\sqrt{x}-1)$
5. $x+\sqrt{x}-2=(x-\sqrt{x})+(2\sqrt{x}-2)$
$=\sqrt{x}(\sqrt{x}-1)+2(\sqrt{x}-1)=(\sqrt{x}-1)(\sqrt{x}+2)$
6. $x-5\sqrt{x}+6=(x-2\sqrt{x})-(3\sqrt{x}-6)=\sqrt{x}(\sqrt{x}-2)-3(\sqrt{x}-2)=(\sqrt{x}-2)(\sqrt{x}-3)$
7. $x\sqrt{x}-1=(\sqrt{x})^3-1^3=(\sqrt{x}-1)(x+\sqrt{x}+1)$
8. $x\sqrt{x}-x+\sqrt{x}-1=x(\sqrt{x}-1)+(\sqrt{x}-1)=(\sqrt{x}-1)(x+1)$
9. $x+2\sqrt{x}-15=(x-3\sqrt{x})+(5\sqrt{x}-15)=\sqrt{x}(\sqrt{x}-3)+5(\sqrt{x}-3)=(\sqrt{x}-3)(\sqrt{x}+5)$
10. $x-2\sqrt{x}-3=(x+\sqrt{x})-(3\sqrt{x}+3)=\sqrt{x}(\sqrt{x}+1)-3(\sqrt{x}+1)=(\sqrt{x}+1)(\sqrt{x}-3)$
\(x+\sqrt{x}=\sqrt{x}\left(\sqrt{x}+1\right)\\ x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)\\ a+3\sqrt{a}-10=a+5\sqrt{a}-2\sqrt{a}-10=\sqrt{a}\left(\sqrt{a}+5\right)-2\left(\sqrt{a}+5\right)=\left(\sqrt{a}-2\right)\left(\sqrt{a}+5\right)\)
\(x\sqrt{x}+\sqrt{x}-x-1=\left(x\sqrt{x}-x\right)+\left(\sqrt{x}-1\right)=x\left(\sqrt{x}-1\right)+\sqrt{x}-1=\left(\sqrt{x}-1\right)\left(x+1\right)\\ x+\sqrt{x}-2=x+2\sqrt{x}-\sqrt{x}-2=\sqrt{x}\left(\sqrt{x}+2\right)-\left(\sqrt{x}+2\right)=\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\\ x-5\sqrt{x}+6=x-2\sqrt{x}-3\sqrt{x}-6=\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)=\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)\)
Mấy bạn còn lại tương tự những bài trên nhé. Nếu còn thắc mắc ở chỗ nào bạn có thể liên hệ mình nhé. Nhớ lần sau bạn tách ra nha, chứ nhiều câu quá.
Bài 1: Rút gọn biểu thức dạng chữ:
1) \(A=\dfrac{2\sqrt{x}+13}{x+5\sqrt{x}+6}+\dfrac{\sqrt{x}-2}{\sqrt{x}+2}-\dfrac{2\sqrt{x-1}}{\sqrt{x}+3}\) ( với \(x\ge0\))
2) \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{3}{\sqrt{x}-3}\right).\dfrac{\sqrt{3}+3}{x+9}\)( với x\(\ge0,\) x\(\ne9\))
cho biểu thuwcsl A= \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{x-9}\)với x≥0,x≠9
a) chứng minh A=\(\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)
b) tính giá trị của A khi x=36
c) tìm x để A<\(\dfrac{1}{2}\)
a: \(A=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)
\(=\dfrac{3\sqrt{x}}{\sqrt{x}-3}\)