3 nhé bạn!
Trong bài ta có căn thức:\(\sqrt{x-3}\)=> x\(\ge\)3 => thay x=3 vào biểu thức=>
GTNN là 3
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Violympic toán 9
Các câu hỏi tương tự
a, cho x=\(\sqrt{2+\sqrt{3}}\) + \(\sqrt{2-\sqrt{3}}\) và y=\(\sqrt{7-2\sqrt{6}}\)
tính giá trị của biểu thức P=\(\left(x-y\right)^{2020}\)
b, tìm GTNN của B=\(x-\sqrt{x-2020}\)
P = \(\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(1-\frac{\sqrt{x}+1}{x+2\sqrt{x}+1}\right)\)
c) Tìm x để A = \(\frac{1}{P}\) đạt GTNN
Cho \(P=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
a, Rút gọn P
b, Tìm GTNN của P
P=\(\left(\frac{\sqrt{x}}{\sqrt{x}-2}+\frac{4x}{2\sqrt{x}-x}\right):\frac{\sqrt{x}+3}{\sqrt{x}-2}\)
tinh P khi x=11-\(4\sqrt{6}\)
Bài 1: Tính :
Csqrt{frac{3sqrt{3}-4}{2sqrt{3}+1}}-sqrt{frac{sqrt{3}+4}{5-2sqrt{3}}}
Bfrac{1}{sqrt{1}+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+frac{1}{sqrt{3}+sqrt{4}}+....+frac{1}{sqrt{99}+sqrt{100}}
Dsqrt{1+sqrt{3+sqrt{13+4sqrt{3}}}}+sqrt{1-sqrt{3-sqrt{13-4sqrt{3}}}}
Bài 2 : Cho Pleft(frac{1}{sqrt{x}-1}+frac{x-sqrt{x}+6}{x+sqrt{x}-2}right):left(frac{sqrt{x}+1}{sqrt{x}+2}+frac{x-sqrt{x}-2}{x+sqrt{x}+2}right)
a, Rút gọn P
b, Tìm GTNN
c, Tìm x để P.frac{x-1}{x^2+8x} -2
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Bài 1: Tính :
\(C=\sqrt{\frac{3\sqrt{3}-4}{2\sqrt{3}+1}}-\sqrt{\frac{\sqrt{3}+4}{5-2\sqrt{3}}}\)
\(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{4}}+....+\frac{1}{\sqrt{99}+\sqrt{100}}\)
\(D=\sqrt{1+\sqrt{3+\sqrt{13+4\sqrt{3}}}}+\sqrt{1-\sqrt{3-\sqrt{13-4\sqrt{3}}}}\)
Bài 2 : Cho \(P=\left(\frac{1}{\sqrt{x}-1}+\frac{x-\sqrt{x}+6}{x+\sqrt{x}-2}\right):\left(\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{x-\sqrt{x}-2}{x+\sqrt{x}+2}\right)\)
a, Rút gọn P
b, Tìm GTNN
c, Tìm x để \(P.\frac{x-1}{x^2+8x}< -2\)
Cho biểu thức: \(B=\left(1-\frac{\sqrt{x}}{1+\sqrt{x}}\right):\left(\frac{\sqrt{x}+3}{\sqrt{x}-2}+\frac{\sqrt{x}+2}{3-\sqrt{x}}+\frac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)\) với \(x\ge0;x\ne4;9\)
a, Rút gọn biểu thức B
b, Tìm x để B < 0
c, Tìm GTNN của B
tìm GTNN của:
\(\frac{5x-17\sqrt{x}+6}{x-2\sqrt{x}-3}\)
Giải hệ pt
1/left{{}begin{matrix}4xsqrt{y+1}+8xleft(4x^2-4x-3right)sqrt{x+1}dfrac{x}{x+1}+x^2left(y+2right)sqrt{left(x+1right)left(y+1right)}end{matrix}right.
2/left{{}begin{matrix}xsqrt{y^2+6}+ysqrt{x^2+3}7xyxsqrt{x^2+3}+ysqrt{y^2+6}x^2+y^2+2end{matrix}right.left{{}begin{matrix}xsqrt{y^2+6}+ysqrt{x^2+3}7xyxsqrt{x^2+3}+ysqrt{y^2+6}x^2+y^2+2end{matrix}right.
3/left{{}begin{matrix}left(2x+y-1right)left(sqrt{x+3}+sqrt{xy}+sqrt{x}right)8sqrt{x}left(sqrt{x+3}+sqrt{xy}right)^2+xy2xleft(6-xright)end...
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Giải hệ pt
1/\(\left\{{}\begin{matrix}4x\sqrt{y+1}+8x=\left(4x^2-4x-3\right)\sqrt{x+1}\\\dfrac{x}{x+1}+x^2=\left(y+2\right)\sqrt{\left(x+1\right)\left(y+1\right)}\end{matrix}\right.\)
2/\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)\(\left\{{}\begin{matrix}x\sqrt{y^2+6}+y\sqrt{x^2+3}=7xy\\x\sqrt{x^2+3}+y\sqrt{y^2+6}=x^2+y^2+2\end{matrix}\right.\)
3/\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\left(2x+y-1\right)\left(\sqrt{x+3}+\sqrt{xy}+\sqrt{x}\right)=8\sqrt{x}\\\left(\sqrt{x+3}+\sqrt{xy}\right)^2+xy=2x\left(6-x\right)\end{matrix}\right.\)
4/\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)\(\left\{{}\begin{matrix}\sqrt{xy+x+2}+\sqrt{x^2+x}-4\sqrt{x}=0\\xy+x^2+2=x\left(\sqrt{xy+2}+3\right)\end{matrix}\right.\)
m.n giúp e mấy bài này vs ạ!!
Giải các phương trình sau:
1.
a. sqrt{x+3}-sqrt{x-4}1
b. sqrt{10-x}+sqrt{x+3}5
c. sqrt{15-x}+sqrt{3-x}6
d. sqrt{x-1}+sqrt{x+1}2
e. sqrt{4x+1}-sqrt{3x+4}1
f. sqrt{x-2sqrt{x-1}}-sqrt{x-1}1
g. sqrt{x+sqrt{2x+1}}+sqrt{x-sqrt{2x-1}}sqrt{2}
h. sqrt{x+sqrt{6x-9}}+sqrt{x-sqrt{6x-9}}sqrt{6}
i. sqrt{x^2-4x+4}+sqrt{x^2-6x+9}1
k. sqrt{x+4-4sqrt{x}}+sqrt{x+9-6sqrt{x}}1
l. sqrt{x+6-4sqrt{x+2}}+sqrt{x+11-6sqrt{x+2}}1
m. sqrt{x+2-4sqrt{x-2}}+sqrt{x+7-6sqrt{x-2}1}
n. sqrt{x}+sqrt{x+sqrt{1-x}}1
o...
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Giải các phương trình sau:
1.
a. \(\sqrt{x+3}-\sqrt{x-4}=1\)
b. \(\sqrt{10-x}+\sqrt{x+3}=5\)
c. \(\sqrt{15-x}+\sqrt{3-x}=6\)
d. \(\sqrt{x-1}+\sqrt{x+1}=2\)
e. \(\sqrt{4x+1}-\sqrt{3x+4}=1\)
f. \(\sqrt{x-2\sqrt{x-1}}-\sqrt{x-1}=1\)
g. \(\sqrt{x+\sqrt{2x+1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
h. \(\sqrt{x+\sqrt{6x-9}}+\sqrt{x-\sqrt{6x-9}}=\sqrt{6}\)
i. \(\sqrt{x^2-4x+4}+\sqrt{x^2-6x+9}=1\)
k. \(\sqrt{x+4-4\sqrt{x}}+\sqrt{x+9-6\sqrt{x}}=1\)
l. \(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\)
m. \(\sqrt{x+2-4\sqrt{x-2}}+\sqrt{x+7-6\sqrt{x-2}=1}\)
n. \(\sqrt{x}+\sqrt{x+\sqrt{1-x}}=1\)
o. \(\sqrt{1-\sqrt{x^2-x}}=\sqrt{x}-1\)
p. \(\sqrt{x^2+6}=x-2\sqrt{x^2-1}\)
q. \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
r. \(\sqrt{x-7}+\sqrt{9-x}=x^2-16x+66\)
s. \(\sqrt{2x-1}+\sqrt{x-2}=\sqrt{x+1}\)
t. \(\sqrt{3x+15}-\sqrt{4x-17}=\sqrt{x+2}\)
u. \(\sqrt{x-1}+\sqrt{x+3}+2\sqrt{\left(x-1\right)\left(x^2-3x+5\right)}=4-2x\)
v. \(\sqrt{x+1}+\sqrt{x+10}=\sqrt{x+2}+\sqrt{x+5}\)
w. \(\sqrt{2x+3+\sqrt{x+2}}+\sqrt{2x+2-\sqrt{x+2}}=1+2\sqrt{x+2}\)
x. \(\sqrt{2x^2-9x+4}+3\sqrt{2x-1}=\sqrt{2x^2+21x-11}\)
y. \(\sqrt{1-x}+\sqrt{x^2-3x+2}+\left(x-2\right)\sqrt{\dfrac{x-1}{x-2}}=3\)
z. \(\left(x-2\right)\left(x+2\right)+4\left(x-2\right)\sqrt{\dfrac{x+2}{x-2}}=-3\)
2.
a. \(\dfrac{2+\sqrt{x}}{\sqrt{2}+\sqrt{2+\sqrt{x}}}+\dfrac{2-\sqrt{x}}{\sqrt{2}-\sqrt{2-\sqrt{x}}}=\sqrt{2}\)
b. \(\dfrac{x}{2+\dfrac{x}{2+\dfrac{x}{2+\dfrac{...}{2+\dfrac{x}{1+\sqrt{1+x}}}}}}=8\) (vế trái có 100 dấu phân thức)
c. \(\sqrt[3]{x+1}+\sqrt[3]{7-x}=2\)
d. \(\sqrt[4]{1-x}+\sqrt[4]{2-x}=\sqrt[4]{3-2x}\)
e. \(\sqrt[4]{1-x^2}+\sqrt[4]{1+x}+\sqrt[4]{1-x}=3\)
f. \(\dfrac{\sqrt[3]{7-x}-\sqrt[3]{x-5}}{\sqrt[3]{7-x}+\sqrt[3]{x-5}}=6-x\)
g. \(\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}=0\)
h. \(\sqrt[3]{\left(x+1\right)^2}+\sqrt[3]{\left(x-1\right)^2}+\sqrt[3]{x^2-1}=1\)
i. \(\sqrt[3]{x+1}+\sqrt[3]{x-1}=\sqrt[3]{5x}\)
k. \(\sqrt[3]{x-2}+\sqrt{x+1}=3\)
l. \(\sqrt[3]{24+x}+\sqrt{12-x}=6\)
m. \(\sqrt[3]{2-x}+\sqrt{x-1}=1\)
n. \(1+\sqrt[3]{x-16}=\sqrt[3]{x+3}\)
o. \(\sqrt[3]{25+x}+\sqrt[3]{3-x}=4\)
p. \(\sqrt[3]{x+3}-\sqrt[3]{6-x}=1\)
Làm nhanh giúp mk nhé mn ơi