RÚT GỌN CÁC BIỂU THỨC SAU:
a) \(A=\sqrt{x-6\sqrt{x}+9}-\sqrt{4x+4\sqrt{x}+1}\)
b) \(B=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
BẠN NÀO LM ĐÚNG MK TIK CHO NHÉ, CAOMON TRC
rút gọn các biểu thức sau:
a) \(A=\sqrt{x^2+8x+16}+\sqrt{x^2-8x+16}Với-4\le x\le4\)\(x\le4\)
b) \(B=\sqrt{9x^2-6x+1}+\sqrt{4x^2-12x+9}\)
c) \(C=\sqrt{x-6\sqrt{x}+9}-\sqrt{4x+4\sqrt{x}+1}\)
d) \(D=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
BẠN NÀO LÀM ĐÚNG MK TIK NHÉ, CAMON TRC
điều kiện -4<=x<=4x<=4
\(a,\sqrt{\left(x+4\right)^2}+\sqrt{\left(x-4\right)^2}\)
\(A=\left|x+4\right|+\left|x-4\right|\)
KẾT HỢP ĐIỀU KIỆN
\(A=x+4+4-x\)
\(A=8\)
\(B=\sqrt{\left(3x\right)^2-6x+1}+\sqrt{\left(2x\right)^2-12x+3^2}\)
\(B=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(2x-3\right)^2}\)
\(B=\left|3x-1\right|+\left|2x-3\right|\)
\(TH1:x>=\frac{3}{2}\)
\(B=3x-1+2x-3\)
\(B=5x-4\)
\(TH2:\frac{1}{3}< =x< \frac{3}{2}\)
\(B=3x-1-2x+3\)
\(B=x+2\)
\(TH3:x< \frac{1}{3}\)
\(B=-3x+1-2x+3\)
\(B=4-5x\)
câu c và câu d tương tự
câu c tách ra: \(C=\sqrt{\left(\sqrt{x}-3\right)^2}-\sqrt{\left(2\sqrt{x}+1\right)^2}\)
còn câu d tách ra :\(D=\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
\(D=\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}\)
bạn tự làm nốt câu c, d nha
rút gọn các biểu thức:
a) \(A=\sqrt{x^2+8x+16}+\sqrt{x^2-8x+16}Với\)VỚI \(-4\le x\le4\)
b) \(B=\sqrt{9x^2-6x+1}+\sqrt{4x^2-12x+9}\)
AI LM ĐÚNG MK TIK CHO NHÉ, CAMON TRC
a, Với \(-4\le x\le4\)
\(A=\sqrt{x^2+8x+16}+\sqrt{x^2-8x+16}\)
\(=\sqrt{\left(x+4\right)^2}+\sqrt{\left(x-4\right)^2}=\left|x+4\right|+\left|x-4\right|\)
b, \(B=\sqrt{9x^2-6x+1}+\sqrt{4x^2-12x+9}\)
\(=\sqrt{\left(3x\right)^2-2.3x+1}+\sqrt{\left(2x\right)^2-2.2x.3x+3^2}\)
\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(2x-3\right)^2}=\left|3x-1\right|+\left|2x-3\right|\)
Rút gọn các biểu thức sau:
A=\(\dfrac{2x+2}{\sqrt{x}}+\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\)
B=\(\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\)
C=\(\left(\dfrac{x+\sqrt{x}-1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}-1}\)
\(A=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\left(x>0,x\ne1\right)\)
\(=\dfrac{2x+2}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2x+2}{\sqrt{x}}+\dfrac{x+\sqrt{x}+1}{\sqrt{x}}-\dfrac{x-\sqrt{x}+1}{\sqrt{x}}=\dfrac{2x+2\sqrt{x}+2}{\sqrt{x}}\)
\(=\dfrac{2\left(x+\sqrt{x}+1\right)}{\sqrt{x}}\)
\(B=\dfrac{2\sqrt{x}-9}{x-5\sqrt{x}+6}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4;9\right)\)
\(=\dfrac{2\sqrt{x}-9}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\dfrac{2\sqrt{x}-9-\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\)
\(C=\left(\dfrac{x+\sqrt{x}-1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\right):\dfrac{1}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)
\(=\left(\dfrac{x+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\right).\left(\sqrt{x}-1\right)\)
\(=\dfrac{x+\sqrt{x}-1-\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)=\dfrac{3\sqrt{x}-2}{x+\sqrt{x}+1}\)
Cho biểu thức:
\(P=\dfrac{x-\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3};x\ge0,x\ne9\)
1) Rút gọn biểu thức P.
2) Tính giá trị của P trong các trường hợp sau:
a) \(x=\dfrac{9}{4}\)
b) \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
3) Tìm x để \(\dfrac{1}{P}>\dfrac{5}{4}\)
1: Ta có: \(P=\dfrac{x-\sqrt{x}}{x-9}+\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{\sqrt{x}-3}\)
\(=\dfrac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{x-\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+3}\)
2)
a) Thay \(x=\dfrac{9}{4}\) vào P, ta được:
\(P=\left(\dfrac{3}{2}+2\right):\left(\dfrac{3}{2}+3\right)=\dfrac{7}{2}:\dfrac{11}{2}=\dfrac{7}{11}\)
b) Ta có: \(x=\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)
\(=5+\sqrt{2}-4-\sqrt{2}\)
=1
Thay x=1 vào P, ta được:
\(P=\dfrac{1+2}{1+3}=\dfrac{3}{4}\)
`Cho biểu thức P=\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right)\div\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
a)Rút gọn P
b)Tìm giá trị nhỏ nhất
Lm nhanh giúp mk nhé!
a) ĐK:\(x\ge0;x\ne9\)
\(P=\left[\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]:\dfrac{2\sqrt{x}-2-\left(\sqrt{x}-3\right)}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}.\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{-3}{\sqrt{x}+3}\)
b)\(P=-\dfrac{3}{\sqrt{x}+3}\)
Có \(\sqrt{x}+3\ge3;\forall x\ge0\)
\(\Leftrightarrow-\dfrac{3}{\sqrt{x}+3}\ge-\dfrac{1}{3}\)
\(P_{min}=-\dfrac{1}{3}\Leftrightarrow x=0\)
a) Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
\(=\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}\)
\(=\dfrac{-3}{\sqrt{x}+3}\)
các bạn giải chi tiết giúp mk nhé. Cảm ơn
1. a> Rút gọn biểu thức sau : A= \(5\left(\frac{1}{\sqrt{2-\sqrt{3}}}+\sqrt{3-\sqrt{5}}-\frac{\sqrt{10}}{2}\right)^2\)+ \(\left(\frac{1}{\sqrt{2+\sqrt{3}}}+\sqrt{3-\sqrt{5}}-\frac{\sqrt{6}}{2}\right)^2\)
b) Cho biểu thức B= \(\left(\frac{\sqrt{x}+1}{\sqrt{x}-1}-\frac{\sqrt{x}-1}{\sqrt{x+1}}-\frac{8\sqrt{x}}{x-1}\right):\left(\frac{\sqrt{x}-x-3}{x-1}-\frac{1}{\sqrt{x}-1}\right)\)
Rút gọn biểu thức B và chứng minh B nhỏ hơn hoặc bằng 1 với mọi x lớn hơn hoặc bằng 0 và x khác 1
Rút gọn các biểu thức sau:
a) $\sqrt{9a^4}$
b) 2$\sqrt{a^{2}}$- 5a (với a<0)
c) $\sqrt{16(1+4x+4x^2)}$ với x $\geq$ $\frac{1}{2}$
d) $\frac{1}{a-3}$$\sqrt{9(a^2-3a+9)}$ với a<3
a) \(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}=\left|3a^2\right|=3a^2\)
b) \(2\sqrt{a^2}-5a=2\left|a\right|-5a=-2a-5a=-7a\)
c) \(\sqrt{16\left(1+4x+4x^2\right)}=\sqrt{\left[4\left(1+2x\right)\right]^2}=\left|4\left(1+2x\right)\right|=4\left(1+2x\right)\)
Rút gọn các biểu thức sau:
a) \(\dfrac{4}{\sqrt{11}-3}-\dfrac{5}{4+\sqrt{11}}\)
b) \(\left(\dfrac{3\sqrt{x}}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}+13}{x+6\sqrt{x}+9}\) với x>0;x\(\ne\)4
a: \(=6+2\sqrt{11}-4+\sqrt{11}=2+3\sqrt{11}\)
b: \(=\dfrac{3x+9\sqrt{x}-2x+4\sqrt{x}}{\left(\sqrt{x}+3\right)\left(x-2\sqrt{x}\right)}\cdot\dfrac{\left(\sqrt{x}+3\right)^2}{\sqrt{x}+13}=\dfrac{\sqrt{x}+3}{x-2\sqrt{x}}\)
Rút gọn các biểu thức sau:
a) A = \(\left(\dfrac{\sqrt{x}}{x-4}+\dfrac{2}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right):\left(\sqrt{x}-2+\dfrac{10-x}{\sqrt{x}+2}\right)\)
b) B = \(\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right):\left(x-y\right)+\dfrac{2\sqrt{y}}{\sqrt{x}+\sqrt{y}}\)
c) C = \(\left(1-\dfrac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}+\dfrac{2+\sqrt{x}}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\)
d) D = \(\sqrt{\dfrac{a+x^2}{x}-2\sqrt{a}}-\sqrt{\dfrac{a+x^2}{x}+2\sqrt{a}}\) với a > 0, x > 0.