BT1: Tính
a, \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}\)
b, \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)
BT2: Rút gọn
\(3x-\sqrt{27}+\frac{\sqrt{x^3+3x^2}}{\sqrt{x+3}}\) ( x ≥ 0 )
BT1: Tính
a, \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{12}\)
b, \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{7}\)
BT2: Rút gọn
\(3x-\sqrt{27}+\frac{\sqrt{x^3+3x^2}}{\sqrt{x+3}}\) ( x ≥ 0 )
rút gọn các biểu thức sau với x ≥ 0
a) \(2\sqrt{3x}-4\sqrt{3x}+27-3\sqrt{3x}\)
b) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+28\)
\(a,=27-5\sqrt{3x}\\ b,=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28=14\sqrt{2x}+28\)
Rút gọn: (Giải chi tiết từng bước)
9) \(2\sqrt{8\sqrt{3}-2\sqrt{5\sqrt{3}}}-3\sqrt{20\sqrt{3}}\)
10) \(\sqrt{12x}-\sqrt{48x}-3\sqrt{3x}+27\) với x \(\ge\) 0
11) \(\sqrt{18x}-5\sqrt{8x}+7\sqrt{18x}+28\) với \(x\ge0\)
12) \(\sqrt{45a}-\sqrt{20a}+4\sqrt{45a}+\sqrt{a}\) với \(a\ge0\)
Cần gấp ạ
9) Sửa: \(2\sqrt{8\sqrt{3}}-2\sqrt{5\text{ }\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=2\sqrt{2^2\cdot2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{2^2\cdot5\sqrt{3}}\)
\(=2\cdot2\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\cdot2\sqrt{5\sqrt{3}}\)
\(=4\sqrt{2\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}\)
\(=4\sqrt{2\sqrt{3}}-8\sqrt{5\sqrt{3}}\)
10) \(\sqrt{12x}-\sqrt{48x}-3\sqrt{3x}+27\)
\(=\sqrt{2^2\cdot3x}-\sqrt{4^2\cdot3x}-3\sqrt{3x}+27\)
\(=2\sqrt{3x}-4\sqrt{3x}-3\sqrt{3x}+27\)
\(=-5\sqrt{3x}++27\)
11) \(\sqrt{18x}-5\sqrt{8x}+7\sqrt{18x}+28\)
\(=\sqrt{3^2\cdot2x}-5\sqrt{2^2\cdot2x}+7\sqrt{3^2\cdot2x}+28\)
\(=3\sqrt{2x}-5\cdot2\sqrt{2x}+7\cdot3\sqrt{2x}+28\)
\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28\)
\(=14\sqrt{2x}+28\)
12) \(\sqrt{45a}-\sqrt{20a}+4\sqrt{45a}+\sqrt{a}\)
\(=\sqrt{3^2\cdot5a}-\sqrt{2^2\cdot5a}+4\sqrt{3^2\cdot5a}+\sqrt{a}\)
\(=3\sqrt{5a}-2\sqrt{5a}+4\cdot3\sqrt{5a}+\sqrt{a}\)
\(=3\sqrt{5a}-2\sqrt{5a}+12\sqrt{5a}+\sqrt{a}\)
\(=13\sqrt{5a}+\sqrt{a}\)
Bài 46 (trang 27 SGK Toán 9 Tập 1)
Rút gọn các biểu thức sau với $x \ge 0$:
a) $2 \sqrt{3x}-4 \sqrt{3x}+27-3 \sqrt{3 x}$ ; b) $3 \sqrt{2 x}-5 \sqrt{8 x}+7 \sqrt{18 x}+28$.
Rút gọn các biểu thức sau với x≥0x≥0:
a) 2\(\sqrt{3x}\)-4\(\sqrt{3x}\)+27-3\(\sqrt{3x}\)=27-5\(\sqrt{3x}\)
b)3\(\sqrt{2x}\)-5\(\sqrt{8x}\)+7\(\sqrt{18x}\)+28
=3\(\sqrt{2x}\)-10\(\sqrt{2x}\)+21\(\sqrt{2x}\)+28
=14\(\sqrt{2x}\)+28=14(\(\sqrt{2x}\)+2)
a) \(2\sqrt{3x}-4\sqrt{3x}+27-3\sqrt{3x}\)
\(=\left(2\sqrt{3x}-4\sqrt{3x}-3\sqrt{3x}\right)+27\)
\(=-5\sqrt{3x}+27\)
b) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+28\)
\(=3\sqrt{2x}-5\sqrt{4.2x}+7\sqrt{9.2x}+28\)
\(=3\sqrt{2x}-5\sqrt{2^2.2x}+7\sqrt{3^2.2x}+28\)
\(=3\sqrt{2x}-5.2\sqrt{2x}+7.3\sqrt{2x}+28\)
\(=\left(3\sqrt{2x}-5.2\sqrt{2x}+7.3\sqrt{2x}\right)+28\)
\(=\left(3-10+21\right)\sqrt{2x}+28\)
\(=14\sqrt{2x}+28\)
Rút gọn các biểu thức sau với x>= 0:
a)\(2\sqrt{3x}-4\sqrt{3x}+27-3\sqrt{3x}\)
b)\(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+28\)
Rút gọn các biểu thức sau:
a) \(A=3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+30\), \(x\ge0\)
b) \(B=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\sqrt{\dfrac{9x^3}{64}}\), \(x>0\)
c) \(C=\dfrac{y}{2}+\dfrac{3}{4}\sqrt{1+9y^2-6y}-\dfrac{3}{2}\), \(y\le\dfrac{1}{3}\)
a) Ta có: \(A=3\sqrt{2x}-5\sqrt{8x}+7\sqrt{18x}+30\)
\(=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+30\)
\(=14\sqrt{2x}+30\)
b) Ta có: \(B=4\sqrt{\dfrac{25x}{4}}-\dfrac{8}{3}\sqrt{\dfrac{9x}{4}}-\dfrac{4}{3x}\cdot\sqrt{\dfrac{9x^3}{64}}\)
\(=4\cdot\dfrac{5\sqrt{x}}{2}-\dfrac{8}{3}\cdot\dfrac{3\sqrt{x}}{2}-\dfrac{4}{3x}\cdot\dfrac{3x\sqrt{x}}{8}\)
\(=10\sqrt{x}-4\sqrt{x}-\dfrac{1}{2}\sqrt{x}\)
\(=\dfrac{11}{2}\sqrt{x}\)
c) Ta có: \(\dfrac{y}{2}+\dfrac{3}{4}\sqrt{9y^2-6y+1}-\dfrac{3}{2}\)
\(=\dfrac{1}{2}y+\dfrac{3}{4}\left(1-3y\right)-\dfrac{3}{2}\)
\(=\dfrac{1}{2}y+\dfrac{3}{4}-\dfrac{9}{4}y-\dfrac{3}{2}\)
\(=-\dfrac{7}{4}y-\dfrac{3}{4}\)
1) Rút gọn biểu thức:
a, \(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)
b, \(\sqrt{4-\sqrt{7}}+\sqrt{4+\sqrt{7}}\)
2) Giải phương trình:
a, \(\left(x\sqrt{\frac{6}{x}}+\sqrt{\frac{2x}{3}}+\sqrt{6x}\right).\sqrt{6x}=2\)
b, \(\left(\sqrt{\frac{3}{x}}+\sqrt{\frac{x}{3}}+\sqrt{3x}\right).\sqrt{3x}=3\)
c, \(\sqrt{x^2+2x+1}-\sqrt{x^2-1}=0\)
d, \(\sqrt{x}+\sqrt{x+1}=\frac{1}{\sqrt{x}}\)
B1: tính : A = \(\sqrt{7+4\sqrt{3}}\) + \(\sqrt{7-4\sqrt{3}}\)
B2: cho P= 3x-\(\sqrt{x^2-10x+25}\)
a, rút gọn P
b, tính P khi x=2
B3: rút gọn : M = \(\dfrac{\sqrt{x^2-2x+1}}{x-1}\)với x khác 1
giúp em zới ạ em cảm mơn nhìu nhìu
\(1.\\ A=\sqrt{\left(2+\sqrt{3}\right)^2}+\sqrt{\left(2-\sqrt{3}\right)^2}\\ =\left|2+\sqrt{3}\right|+\left|2-\sqrt{3}\right|\\ =2+\sqrt{3}+2-\sqrt{3}=4\)
\(2.\\a.\\ P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\\ b.\\ x=2\Rightarrow P=3\)
\(3.\\ M=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)
\(\cdot x>1\Rightarrow M=1\\ \cdot x=1\Rightarrow M=0\\\cdot x< 1\Rightarrow M=-1\)
B1.
Ta có:A\(=\sqrt{3+4\sqrt{3}+4}+\sqrt{3-4\sqrt{3}+4}\)
\(=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\)
\(=\sqrt{3}+2+\sqrt{3}-2=2\sqrt{3}\)
Bài 1 :
\(A=\sqrt{\left(\sqrt{3}+2\right)^2}+\sqrt{\left(\sqrt{3}-2\right)^2}\\ =\sqrt{3}+2+2-\sqrt{3}=4\)
Bài 2 :
a) \(P=3x-\sqrt{\left(x-5\right)^2}=3x-\left|x-5\right|\)
b) khi x = 2 thì \(P=3.2-\left|2-5\right|=3\)
Bài 3 :
\(M=\dfrac{\sqrt{\left(\sqrt{x}-1\right)^2}}{x-1}=\dfrac{\left|\sqrt{x}-1\right|}{x-1}\)
GIÚP EM ĐI Ạ
TÍNH:
\(\frac{3-\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}{3-\sqrt{3+\sqrt{6+\sqrt{3}}}}+\frac{2+\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}{3+\sqrt{6+\sqrt{3+\sqrt{6+\sqrt{3}}}}}\)
\(\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1+\frac{\sqrt{3}}{2}}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{1-\frac{\sqrt{3}}{2}}}\)
\(\frac{1}{\sqrt{\frac{5}{13}}+\sqrt{\frac{5}{7}}+1}+\frac{1}{\sqrt{\frac{7}{5}}+\sqrt{\frac{7}{13}}+1}+\frac{1}{\sqrt{1\frac{6}{7}}+1+\sqrt{2\frac{3}{5}}}\)
RÚT GỌN
\(\sqrt{\left(x-1\right)^2}-x\) với x lớn hơn 1
GIẢI PHƯƠNG TRÌNH
\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)
Bài rút gọn
\(\sqrt{\left(x-1\right)^2}-x=\left|x-1\right|-x\)
\(=\left(x-1\right)-x=x-1-x=-1\left(x>1\right)\)
Bài gpt:
\(\sqrt{x^2-3x+2}+\sqrt{x^2-4x+3}=0\)
Đk:\(-1\le x\le3\)
\(pt\Leftrightarrow\sqrt{\left(x-1\right)\left(x-2\right)}+\sqrt{\left(x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x-2}+\sqrt{x-3}\right)=0\)
Dễ thấy:\(\sqrt{x-2}+\sqrt{x-3}=0\) vô nghiệm
Nên \(\sqrt{x-1}=0\Rightarrow x-1=0\Rightarrow x=1\)