\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}-2\left(x\ge0\right)\)
Những câu hỏi liên quan
Bài 1: Cho \(A=\left(\dfrac{x-4}{\sqrt{x}-2}+\dfrac{x\sqrt{x}-8}{4-x}\right):\left[\dfrac{\left(\sqrt{x}-2\right)^2+2\sqrt{x}}{\sqrt{x}+2}\right]\)với \(x\ge0\); \(x\ne4\)
a, Rút gọn A
b, CMR: \(A< 1\) với \(x\ge0\); \(x\ne4\)
c, Tìm x để A nguyên
a: \(A=\left(\dfrac{\left(x-4\right)\left(\sqrt{x}+2\right)-x\sqrt{x}+8}{x-4}\right):\dfrac{x-2\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8-x\sqrt{x}+8}{x-4}\cdot\dfrac{\sqrt{x}+2}{x-2\sqrt{x}+4}\)
\(=\dfrac{2x-4\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{1}{x-2\sqrt{x}+4}=\dfrac{2\sqrt{x}}{x-2\sqrt{x}+4}\)
b: \(A-1=\dfrac{2\sqrt{x}-x+2\sqrt{x}-4}{x-2\sqrt{x}+4}\)
\(=\dfrac{-x+4\sqrt{x}-4}{x-2\sqrt{x}+4}=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-1\right)^2+3}< 0\)
=>A<1
c: \(2\sqrt{x}>=0;x-2\sqrt{x}+4=\left(\sqrt{x}-1\right)^2+3>0\)
=>A>=0 với mọi x thỏa mãn ĐKXĐ
mà A<1
nên 0<=A<1
=>Để A nguyên thì A=0
=>x=0
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Rút gọn các biểu thức sau:
\(C=\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right).\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}-2}\)
(với \(x\ge0,x\ne4,x\ne9\))
\(D=\left(\dfrac{\sqrt{x}+2}{x-9}-\dfrac{\sqrt{x}-2}{x+6\sqrt{x}+9}\right).\dfrac{x\sqrt{x}+3x-9\sqrt{x}-27}{\sqrt{x}-2}\)
(với \(x\ge0,x\ne4,x\ne9\))
\(C=\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right).\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}-2}\)
\(=\left[\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{x\left(\sqrt{x}+2\right)-4\left(\sqrt{x}+2\right)}{\sqrt{x}-2}\)
\(=\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right].\dfrac{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)
\(=\left[\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}-\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right].\left(\sqrt{x}+2\right)^2\)
\(=\dfrac{6\sqrt{x}}{\sqrt{x}-2}\)
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\(C=\left[\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{\sqrt{x}\left(x-4\right)+2\left(x-4\right)}{\sqrt{x}-2}\) (\(x\ge0,x\ne4,x\ne9\))
\(C=\left[\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}\right].\dfrac{\left(\sqrt{x}+2\right)\left(x-4\right)}{\sqrt{x}-2}\)
\(C=\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}{\sqrt{x}-2}\)
\(C=\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}.\left(\sqrt{x}+2\right)^2\)
\(C=\dfrac{2}{\sqrt{x}-2}\)
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a)3sqrt{40sqrt{12}}+4sqrt{sqrt{75}}-5sqrt{5sqrt{48}}b)sqrt{8sqrt{3}}+3sqrt{20sqrt{3}}-2sqrt{45sqrt{3}}c)left(sqrt{x}-1right).left(x+sqrt{x}+1right)left(xge0;yge0right)d)left(sqrt{x}+1right)left(x+1-sqrt{x}right)left(xge0;yge0right)e)left(sqrt{x}+yright)left(x+y^2-ysqrt{2}right)left(xge0;yge0right)
Đọc tiếp
a)\(3\sqrt{40\sqrt{12}}+4\sqrt{\sqrt{75}}-5\)\(\sqrt{5\sqrt{48}}\)
b)\(\sqrt{8\sqrt{3}}+3\sqrt{20\sqrt{3}}-2\sqrt{45\sqrt{3}}\)
c)\(\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)\left(x\ge0;y\ge0\right)\)
d)\(\left(\sqrt{x}+1\right)\left(x+1-\sqrt{x}\right)\left(x\ge0;y\ge0\right)\)
e)\(\left(\sqrt{x}+y\right)\left(x+y^2-y\sqrt{2}\right)\left(x\ge0;y\ge0\right)\)
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Rút gọna)3sqrt{40sqrt{12}}+4sqrt{sqrt{75}}-5sqrt{5sqrt{48}}b)sqrt{8sqrt{3}}+3sqrt{20sqrt{3}}-2sqrt{45sqrt{3}}c)left(sqrt{x}-1right).left(x+sqrt{x}+1right)left(xge0;yge0right)d)left(sqrt{x}+1right)left(x+1-sqrt{x}right)left(xge0;yge0right)e)left(sqrt{x}+yright).left(x+y^2-ysqrt{2}right)left(xge0;yge0right)
Đọc tiếp
Rút gọn
a)\(3\sqrt{40\sqrt{12}}+4\sqrt{\sqrt{75}}-5\)\(\sqrt{5\sqrt{48}}\)
b)\(\sqrt{8\sqrt{3}}+3\sqrt{20\sqrt{3}}-2\sqrt{45\sqrt{3}}\)
c)\(\left(\sqrt{x}-1\right).\left(x+\sqrt{x}+1\right)\left(x\ge0;y\ge0\right)\)
d)\(\left(\sqrt{x}+1\right)\left(x+1-\sqrt{x}\right)\left(x\ge0;y\ge0\right)\)
e)\(\left(\sqrt{x}+y\right).\left(x+y^2-y\sqrt{2}\right)\left(x\ge0;y\ge0\right)\)
\(\left(\dfrac{x-2\sqrt{x}}{x-4}-1\right):\left(\dfrac{4-x}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{3-\sqrt{x}}-\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
voi x\(\ge0,x\ne4;9\)
TRẦN MINH HOÀNGKhôi Bùi tran nguyen bao quanMysterious PersoDƯƠNG PHAN KHÁNH DƯƠNGn ghi ket qua thoi cx dc
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\(\left(\dfrac{x-2\sqrt{x}}{x-4}-1\right):\left(\dfrac{4-x}{x-\sqrt{x}-6}-\dfrac{\sqrt{x}-2}{3-\sqrt{x}}-\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\right)=\dfrac{x-2\sqrt{x}-x+4}{x-4}:\left(\dfrac{4-x}{x-\sqrt{x}-6}+\dfrac{\sqrt{x}-2}{\sqrt{x}-3}-\dfrac{\sqrt{x}-3}{\sqrt{x}+2}\right)=\dfrac{-2\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\left[\dfrac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right]=\dfrac{-2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\left[\dfrac{4-x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}+\dfrac{x-4}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}-\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\right]=\dfrac{-2}{\sqrt{x}+2}:\dfrac{4-x+x-4-\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}:\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}:\dfrac{\sqrt{x}+3}{\sqrt{x}+2}=\dfrac{-2}{\sqrt{x}+2}.\dfrac{\sqrt{x}+2}{\sqrt{x}+3}=\dfrac{-2}{\sqrt{x}+3}\)
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Rút gọn : a) dfrac{asqrt{b}-bsqrt{a}}{sqrt{a}-sqrt{b}}-sqrt{ab}
b)dfrac{x+4y-4sqrt{xy}}{sqrt{x}-2sqrt{y}}+dfrac{y+sqrt{xy}}{sqrt{x}+sqrt{y}}left(xge0;yge0;xne4yright)
c)dfrac{x+4sqrt{x}+4}{sqrt{x}+2}+dfrac{4-x}{sqrt{x}-2}left(xge0;xne4right)
d)dfrac{9-x}{sqrt{3x}+3}-dfrac{9-6sqrt{x}+x}{sqrt{x}-3}
e)dfrac{left(sqrt{x}-sqrt{y}right)^2+4sqrt{xy}}{sqrt{x}+sqrt{y}}-dfrac{xsqrt{y}+ysqrt{x}}{sqrt{xy}}
g)left(2-dfrac{a-3sqrt{a}}{sqrt{a}-3}right)left(2-dfrac{5sqrt{a}-sqrt{ab}}{sqrt{b}-5}right)với a,...
Đọc tiếp
Rút gọn : a) \(\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\)
b)\(\dfrac{x+4y-4\sqrt{xy}}{\sqrt{x}-2\sqrt{y}}+\dfrac{y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\left(x\ge0;y\ge0;x\ne4y\right)\)
c)\(\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}+\dfrac{4-x}{\sqrt{x}-2}\left(x\ge0;x\ne4\right)\)
d)\(\dfrac{9-x}{\sqrt{3x}+3}-\dfrac{9-6\sqrt{x}+x}{\sqrt{x}-3}\)
e)\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}\)
g)\(\left(2-\dfrac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\dfrac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)với\) a, b \(\ge\)0 , a \(\ne\)9; b\(\ne\)25
Mọi người giúp tớ với , cảm ơn nhiều nhiều ạ !!
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
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Tất cả đều có điều kiện \(x\ge0\)
a,\(\sqrt{x^2-6x+9}+x=11\)
b,\(\sqrt{3x^2-4x+3=1-2x}\)
c,\(\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\)
d,\(\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\)
Đề yc giải pt à em?
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Câu b bạn có bị lỗi dấu căn không mà sao nó kéo dài cả 2 vế pt vậy :v
\(a,\sqrt{x^2-6x+9}+x=11\\ \Leftrightarrow\sqrt{\left(x-3\right)^2}=11-x\)
\(\Leftrightarrow\left|x-3\right|=11-x\\ TH_1:x\ge3\\ x-3=11-x\\ \Leftrightarrow2x=14\\ \Leftrightarrow x=7\left(tm\right)\)
\(TH_2:x< 3\\ -x+3=11-x\\ \Leftrightarrow-x+x=11-3\\ \Leftrightarrow0=8\left(VL\right)\)
Vậy \(S=\left\{7\right\}\)
\(c,\sqrt{16\left(x+1\right)}-\sqrt{9\left(x+1\right)}=4\) \(\left(dk:x\ge-1\right)\)
\(\Leftrightarrow\sqrt{4^2}.\sqrt{\left(x+1\right)}-\sqrt{3^2}.\sqrt{\left(x+1\right)}=4\left(1\right)\)
Đặt \(a=\sqrt{x+1}\left(a\ge0\right)\)
Pt trở thành : \(4a-3a=4\Leftrightarrow a=4\left(tmdk\right)\)
\(\Rightarrow\sqrt{x+1}=4\\ \Rightarrow\left(\sqrt{x+1}\right)^2=16\\ \Rightarrow\left|x+1\right|=16\)
\(TH_1:x\ge-1\\ x+1=16\Leftrightarrow x=15\left(tm\right)\\ TH_2:x< -1\\ -x-1=16\Leftrightarrow x=-17\left(tm\right)\)
Nhưng loại TH2 vì dk ban đầu là \(x\ge-1\)
Vậy \(S=\left\{15\right\}\)
\(d,\sqrt{9x+9}+\sqrt{4x+4}=\sqrt{x+1}\left(dk:x\ge-1\right)\\ \Leftrightarrow\sqrt{9}.\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}-\sqrt{x+1}=0\)
Đặt \(\sqrt{x+1}=a\left(a\ge0\right)\)
Tới đây bạn làm tương tự câu c nha.
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giúp mình giải bpt vs
\(\dfrac{\left|2x-1\right|-x}{2x}>1;\dfrac{2-\left|x-2\right|}{x^2-1}\ge0;\dfrac{\sqrt{x+4}-2}{4-9x^2}\le0;\dfrac{x^2-2x-3}{\sqrt[3]{3x-1}+\sqrt[3]{4-5x}}\ge0;\)\(3x^2-10x+3\ge0;\left(\sqrt{2}-x\right)\left(x^2-2\right)\left(2x-4\right)< 0;\dfrac{1}{x+9}-\dfrac{1}{x}>\dfrac{1}{2};\dfrac{2}{1-2x}\le\dfrac{3}{x+1}\)
A=\(\left(\frac{3\sqrt{x}}{\sqrt{x}+2}-\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{8\sqrt{x}}{x-4}\right):\left(2-\frac{2\sqrt{x}+3}{\sqrt{x}+2}\right)\left(x\ge0,x\ne4\right)\)
a, Rút gọn A.
b, Tìm GTNN của A khi x>4
\(A=\frac{3\sqrt{x}\left(\sqrt{x}-2\right)-\sqrt{x}\left(\sqrt{x}+2\right)+8\sqrt{x}}{x-4}:\frac{2\left(\sqrt{x}+2\right)-2\sqrt{x}-3}{\sqrt{x}+2}\)
\(A=\frac{2x}{x-4}.\left(\sqrt{x}+2\right)=\frac{2x\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(A=\frac{2x}{\sqrt{x}-2}\)
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