bài 1 : cho biểu thức
\(p=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
a, rút gọn với x>0 ;\(x\ne4;x\ne1\)
b, tìm giá trị của x để Q <0
Bài 2:Cho biểu thức P=\(\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\).\(\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\)
a)Rút gọn BT
b)So sánh P với -\(2\sqrt{x}\)
a) \(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right).\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\left(đk:x>0\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\dfrac{1-x}{2\sqrt{x}}\right)^2=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{x-1}.\dfrac{\left(x-1\right)^2}{4x}=\dfrac{-4\sqrt{x}\left(x-1\right)}{4x}=\dfrac{1-x}{\sqrt{x}}\)
b) \(P-\left(-2\sqrt{x}\right)=\dfrac{1-x}{\sqrt{x}}+2\sqrt{x}=\dfrac{1-x+2x}{\sqrt{x}}=\dfrac{1+x}{\sqrt{x}}>0\)
\(\Rightarrow P>-2\sqrt{x}\)
a, ĐK: \(x\ge0;x\ne1\)
\(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2-\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\left(2-2x\right)^2}{16x}\)
\(=\dfrac{-4\sqrt{x}}{x-1}.\dfrac{4\left(x-1\right)^2}{16x}\)
\(=-\dfrac{x-1}{\sqrt{x}}\)
a: Ta có: \(P=\left(\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\right)\cdot\left(\dfrac{1}{2\sqrt{x}}-\dfrac{\sqrt{x}}{2}\right)^2\)
\(=\dfrac{x-2\sqrt{x}+1-x-2\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)^2\cdot\left(\sqrt{x}+1\right)^2}{4x}\)
\(=\dfrac{-4\sqrt{x}\left(x-1\right)}{4x}\)
\(=\dfrac{-x+1}{\sqrt{x}}\)
Rút gọn biểu thức dạng chữ:
Q=\(\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right).\left(x+\sqrt{x}\right)\) với x ≥0, x ≠1
A= \(A=\left(\dfrac{1}{\sqrt{x}-2}-\dfrac{1}{\sqrt{x}+2}+\dfrac{4\sqrt{x}}{4-x}\right):\dfrac{\sqrt{x}+1}{x-4}\) với x ≥0, x ≠ 4
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\right):\dfrac{1}{x+6\sqrt{x}+9}\) với x ≥ 0, x ≠ 9
Hộ vs ạ
1.
\(Q=\left[\frac{\sqrt{x}+2}{(\sqrt{x}+1)^2}-\frac{\sqrt{x}-2}{(\sqrt{x}-1)(\sqrt{x}+1)}\right].\sqrt{x}(\sqrt{x}+1)\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)}{\sqrt{x}+1}-\frac{\sqrt{x}(\sqrt{x}-2)}{\sqrt{x}-1}\)
\(=\frac{\sqrt{x}(\sqrt{x}+2)(\sqrt{x}-1)-\sqrt{x}(\sqrt{x}-2)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{2x}{x-1}\)
2.
\(A=\left[\frac{\sqrt{x}+2-(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}-\frac{4\sqrt{x}}{x-4}\right].\frac{x-4}{\sqrt{x}+1}\)
\(=\left(\frac{4}{x-4}-\frac{4\sqrt{x}}{x-1}\right).\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{x-4}.\frac{x-4}{\sqrt{x}+1}=\frac{4(1-\sqrt{x})}{\sqrt{x}+1}\)
3.
\(A=\left[\frac{\sqrt{x}(\sqrt{x}-3)+2\sqrt{x}(\sqrt{x}+3)}{(\sqrt{x}-3)(\sqrt{x}+3)}-\frac{3x+9}{(\sqrt{x}-3)(\sqrt{x}+3)}\right]:\frac{1}{(\sqrt{x}+3)^2}\)
\(=\frac{3\sqrt{x}-9}{(\sqrt{x}-3)(\sqrt{x}+3)}.(\sqrt{x}+3)^2=\frac{3(\sqrt{x}-3)}{(\sqrt{x}-3)(\sqrt{x}+3)}(\sqrt{x}+3)^2=3(\sqrt{x}+3)\)
Cho biểu thức P = \(\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) ( với x ≥ 0; x ≠ 1 )
a,Rút gọn biểu thức P
b,Tính giá trị của P khi x = \(6-2\sqrt{5}\)
\(a,P=\dfrac{3\left(x+2\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\left(dk:x\ge0,x\ne1\right)\)
\(=\dfrac{3\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}}{\sqrt{x}-1}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{3\sqrt{x}-\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}-2}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}-1\right)}{\sqrt{x}-1}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\\ =\dfrac{2\left(\sqrt{x}+2\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}+2}\\ =\dfrac{2\sqrt{x}+4-\sqrt{x}-1}{\sqrt{x}+2}\\ =\dfrac{\sqrt{x}+3}{\sqrt{x}+2}\)
\(b,x=6-2\sqrt{5}=\left(\sqrt{5}-1\right)^2\)
\(\Rightarrow P=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}+3}{\sqrt{\left(\sqrt{5}-1\right)^2}+2}=\dfrac{\left|\sqrt{5}-1\right|+3}{\left|\sqrt{5}-1\right|+2}=\dfrac{\sqrt{5}-1+3}{\sqrt{5}-1+2}=\dfrac{\sqrt{5}+2}{\sqrt{5}+1}\)
* Cho biểu thức
A= \(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\) với x > 0, x ≠ 1
a. Rút gọn biểu thức A
b. Tính giá trị của x khi A > \(\dfrac{1}{6}\)
\(A=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right).\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\text{x > 0, x ≠ 1}\)
\(A=\dfrac{\sqrt{x}-\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{x-1-x+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\) \(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}.\dfrac{1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{1}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)^2}\)
Cho biểu thức A= \(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
P = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
a, Rút gọn biểu thức A
b, Tìm các giá trị để \(\dfrac{P}{A}\left(x-1\right)=0\)
\(a,A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(x>0;x\ne1\right)\\ A=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\\ A=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
\(b,\dfrac{P}{A}\left(x-1\right)=0\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}}{\sqrt{x}+1}\cdot\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow x=0\left(\sqrt{x}+1>0\right)\)
a) \(A=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\left(đk:x>0,x\ne1\right)\)
\(=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b) \(\dfrac{P}{A}\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}-1}:\dfrac{\sqrt{x}+1}{\sqrt{x}}.\left(x-1\right)=0\)
\(\Leftrightarrow\dfrac{\sqrt{x}}{\sqrt{x}-1}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow x=0\)( do \(\sqrt{x}+1\ge1>0\))(không thỏa đk)
Vậy \(S=\varnothing\)
Rút gọn các biểu thức sau:
a) \(\left(\dfrac{3-\sqrt{3}}{\sqrt{3}-1}+1\right)\left(\sqrt{3}-1\right)\)
b) \(\left(\dfrac{1}{\sqrt{x}+2}-\dfrac{1}{\sqrt{x}}\right)\left(\dfrac{2}{\sqrt{x+1}}\right)\) với x>0
\(=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=3-1=2\)
b: \(=\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{2}{\sqrt{x}+1}=\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}\)
a, \(=\left(\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}+1\right)\left(\sqrt{3}-1\right)=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)=2\)
b, với x > 0
\(=\left(\dfrac{\sqrt{x}-\sqrt{x}-2}{\sqrt{x}\left(\sqrt{x}+2\right)}\right)\left(\dfrac{2}{\sqrt{x+1}}\right)\)
\(=-\dfrac{-4}{\sqrt{x}\left(\sqrt{x}+2\right)\sqrt{x+1}}=\dfrac{4}{\left(\sqrt{x}+2\right)\sqrt{x^2+x}}\)
cho biểu thức Q=\(\left(\dfrac{1}{\sqrt{X}-1}-\dfrac{1}{\sqrt{X}}\right):\left(\dfrac{\sqrt{X}+1}{\sqrt{X}-2}-\dfrac{\sqrt{X}+2}{\sqrt{X-1}}\right)\)
a rút gọn Q
b tìm x để Q>0
b) Q > 0
⇔ \(\dfrac{\sqrt{\text{x}}-2}{3\sqrt{\text{x}}}\) > 0
Do \(\text{3}\sqrt{\text{x}}>0\) ∀x⩾0
⇒ \(\sqrt{\text{x}}-2>0\)
⇔ \(\sqrt{\text{x}}>2\)
⇔ x > 4
Vậy x > 4 thì Q > 0
Bài 1:Cho biểu thức B= \(\left(\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}}{\sqrt{x}-3}\right):\dfrac{\sqrt{x}}{\sqrt{x}-2}\) (x > 0, x≠ 9)
a) Rút gọn B
c) Giá trị x để B = \(\dfrac{3}{2}\)
Bài 2: Một khu vườn có chuvi = 46 m, nếu tăng chiều dài 5m và giảm chều rộng 3m thì hình chữ nhật mới có chiều dài gấp 4 lần chiều rộng. Tính diện tích khu vườn hình chữ nhật ban đầu
(mink đag cần rất gấp)
Gọi chiều dài, chiều rộng hình chữ nhật lần lượt là x ; y > 0, m
Chu vi hình chữ nhật là : \(P=\left(a+b\right).2=46\)
Nếu tăng chiều dài 5m, giảm chiều rộng 3m thì hình chữ nhật có chiều dài gấp 4 lần chiều rộng : \(a+5=4\left(b-3\right)\)
Ta có hệ phương trình sau : \(\left\{{}\begin{matrix}\left(a+b\right).2=46\\a+5=4\left(b-3\right)\end{matrix}\right.\)
giải hệ ta được a = 15 ; b = 8
Vậy diện tích hình chữ nhật là : \(a.b=15.8=120\)m2
Rút gọn biểu thức:
1, \(B=\left(\dfrac{x.\sqrt{x}+x+\sqrt{x}}{x.\sqrt{x}-1}-\dfrac{\sqrt{x}+3}{1-\sqrt{x}}\right).\dfrac{x-1}{2x+\sqrt{x}-1}\)với x>-0, x khác 1, x khác \(\dfrac{1}{4}\)
2, \(A=\dfrac{\left(\sqrt{x}-1\right)^2.\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{3\sqrt{x}+1}{x-1}\) với x\(\ge\)0:x\(\ne\)0
* Cho biểu thức
A= \(\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right).\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)( với x > 0,x ≠1)
a. Rút gọn biểu thức A
b. Tính giá trị của x khi A=\(\dfrac{1}{2}\)
\(=>A=\left(\dfrac{x-1}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left[\dfrac{\sqrt{x}+1-2}{x-1}\right]\)
\(=>A=\dfrac{\sqrt{x}+1}{\sqrt{x}}.\dfrac{1}{\sqrt{x}+1}=\dfrac{1}{\sqrt{x}}\)
b,\(=>\dfrac{1}{\sqrt{x}}=\dfrac{1}{2}=>\sqrt{x}=2=>x=\sqrt{2}\left(tm\right)\)