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a, \(\sqrt{2}x-\sqrt{50}=0\Leftrightarrow\sqrt{2}x-5\sqrt{2}=0\Leftrightarrow\sqrt{2}\left(x-5\right)=0\Leftrightarrow x=5\)
b, \(\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}\Leftrightarrow\sqrt{3}\left(x+1\right)=5\sqrt{3}\Leftrightarrow x+1=5\Leftrightarrow x=4\)
c, \(\sqrt{3}x^2-\sqrt{12}=0\Leftrightarrow\sqrt{3}\left(x^2-2\right)=0\Leftrightarrow x^2-2=0\Leftrightarrow x=\pm\sqrt{2}\)
d, \(\dfrac{x^2}{\sqrt{5}}-\sqrt{20}=0\Leftrightarrow\dfrac{1}{\sqrt{5}}\left(x^2-10\right)=0\Leftrightarrow x^2-10=0\Leftrightarrow x=\pm\sqrt{10}\)
a) \(\sqrt{2}x-\sqrt{50}=0\)
\(\Leftrightarrow\sqrt{2}x=\sqrt{50}\)
\(\Leftrightarrow x=\frac{\sqrt{50}}{\sqrt{2}}=\sqrt{\frac{50}{2}}=\sqrt{25}=5\)
b) \(\sqrt{3}x+\sqrt{3}=\sqrt{12}+\sqrt{27}\)
\(\Leftrightarrow\sqrt{3}\left(x+1\right)=2\sqrt{3}+3\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}x=5\sqrt{3}\)
\(\Leftrightarrow x=5\)
c) \(\sqrt{3}x^2-\sqrt{12}=0\)
\(\Leftrightarrow\sqrt{3}\left(x^2-2\right)=0\)
\(\Leftrightarrow x^2-2=0\)
\(\Leftrightarrow x^2=2\Leftrightarrow\left[\begin{array}{nghiempt}x=\sqrt{2}\\x=-\sqrt{2}\end{array}\right.\)
d) \(\frac{x^2}{\sqrt{5}}-\sqrt{20}=0\)
\(\Leftrightarrow\)\(\frac{1}{\sqrt{5}}\left(x^2-10\right)=0\)
\(\Leftrightarrow x^2-10=0\)
\(\Leftrightarrow x^2=10\Leftrightarrow\left[\begin{array}{nghiempt}x=\sqrt{10}\\x=-\sqrt{10}\end{array}\right.\)
a) \(\sqrt{3}x-\sqrt{12}=0< =>\sqrt{3}x=\sqrt{12}=>x=2\)
Vay S = { 2 }
b) \(\sqrt{2}x+\sqrt{2}=\sqrt{8}+\sqrt{18}< =>\sqrt{2}x=\sqrt{8}+\sqrt{18}-\sqrt{2}< =>\sqrt{2}x=2\sqrt{2}+3\sqrt{2}-\sqrt{2}\) <=> \(\sqrt{2}x=4\sqrt{2}=>x=4\)
Vay S = { 4 }
c) \(\sqrt{5}x^2-\sqrt{20}=0< =>\sqrt{5}x^2=\sqrt{20}< =>x^2=2=>x=\sqrt{2}\)
Vay S = {\(\sqrt{2}\) }
d) \(\sqrt{x^2+6x+9}=3x+6< =>\sqrt{\left(x+3\right)^2}=3x+6< =>x+3=3x+6< =>-2x=\) \(3=>x=-\dfrac{3}{2}\)
Vay S = { - 3/2 }
e) \(\sqrt{x^2-4x+4}-2x+5=0< =>\sqrt{\left(x-2\right)^2}-2x+5=0< =>x-2-2x+5=0\) <=> \(-x+3=0< =>-x=-3=>x=3\)
Vay S = { 3 }
F) \(\sqrt{\dfrac{2x-3}{x-1}}=2\)
<=> \(\dfrac{2x-3}{x-1}=4< =>2x-3=4x-4< =>-2x=-1=>x=\dfrac{1}{2}\)
Vay S = { 1/2 }
g) \(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2< =>\sqrt{\dfrac{2x-3}{x-1}}=2< =>\dfrac{2x-3}{x-1}=4< =>2x-3=4x-4< =>-2x=-1=>x=\dfrac{1}{2}\)
bạn chưa có ĐKXĐ nên chưa xét kết quả có đúng vs Đk ko, có vài câu sai kết quả
Trần Hữu Ngọc Minh xem tôi làm có đúng ko?
Giải:
a, \(\sqrt{2}.x-\sqrt{50}=0\)
\(\Leftrightarrow\sqrt{2}.x=\sqrt{50}\Leftrightarrow\sqrt{2}.x=\sqrt{25.2}\)
\(\Leftrightarrow\sqrt{2}.x=\sqrt{25}.\sqrt{2}\Leftrightarrow\sqrt{2}.x=5\sqrt{2}\)
\(\Leftrightarrow x=5\)
c, \(\sqrt{3}.x^2-\sqrt{12}=0\)
\(\Leftrightarrow\sqrt{3}.x^2=\sqrt{12}\)
\(\Leftrightarrow\sqrt{3}.x^2=\sqrt{4.3}\)
\(\Leftrightarrow\sqrt{3}.x^2=\sqrt{4}.\sqrt{3}\)
\(\Leftrightarrow\sqrt{3}.x^2=2\sqrt{3}\)
\(\Leftrightarrow x^2=2\)
\(\Leftrightarrow x=\pm\sqrt{2}\)
d, \(\frac{x^2}{\sqrt{5}}-\sqrt{20}=0\)
\(\Leftrightarrow\frac{x^2}{\sqrt{5}}=\sqrt{20}\)
\(\Leftrightarrow x^2=\sqrt{5}.\sqrt{20}\)
\(\Leftrightarrow x^2=\sqrt{100}\)
\(\Leftrightarrow x=\pm10\)
\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(\Rightarrow\sqrt{x}+3\)
\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)
\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)
\(\Rightarrow\sqrt{y}-1\)
\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)
\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)
\(\Rightarrow\sqrt{xy}\)
\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)
\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)
\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)
\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)
\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)
\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)
\(\dfrac{\sqrt{14-6\sqrt{5}}}{\sqrt{5}-3}\)
\(=\dfrac{\sqrt{\left(3-\sqrt{5}\right)^2}}{\sqrt{5}-3}\)
\(=\dfrac{3-\sqrt{5}}{\sqrt{5}-3}\)
= - 1
\(\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{6+2\sqrt{5}}}{2}\)
\(=\dfrac{\sqrt{\left(\sqrt{5}+1\right)^2}}{2}\)
\(=\dfrac{\sqrt{5}+1}{2}\)
\(\dfrac{2+\sqrt{2}}{\sqrt{1,5+\sqrt{2}}}\)
\(=\dfrac{2\sqrt{2}+2}{\sqrt{3+2\sqrt{2}}}\)
\(=\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{\left(\sqrt{2}+1\right)^2}}\)
\(=\dfrac{2\left(\sqrt{2}+1\right)}{\sqrt{2}+1}\)
= 2
\(\dfrac{\sqrt{20}}{\sqrt{5}}+\dfrac{\sqrt{117}}{\sqrt{13}}+\dfrac{\sqrt{272}}{\sqrt{17}}+\dfrac{\sqrt{105}}{\sqrt{2\dfrac{1}{7}}}\)
\(=4+9+16+49\)
= 78
\(\dfrac{x\sqrt{x}-y\sqrt{y}}{x+\sqrt{xy}+y}\)
\(=\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{x+\sqrt{xy}+y}\)
\(=\sqrt{x}-\sqrt{y}\)
\(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{\left(2+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)+\left(2-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}{\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
\(\left[-\text{tử}-\right]=\sqrt{2}\left(2+\sqrt{3}\right)-\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)^2}+\sqrt{2}\left(2-\sqrt{3}\right)+\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)^2}\)
\(=4\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}\)
\(\left[-\text{mẫu}-\right]=2-\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}-\sqrt{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}\)
\(=2-\sqrt{\left(\sqrt{3}-1\right)^2}+\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-3}\)
\(=2-\left(\sqrt{3}-1\right)+\left(\sqrt{3}+1\right)-1\)
= 3
Ta có:
\(\dfrac{4\sqrt{2}-\sqrt{2+\sqrt{3}}+\sqrt{2-\sqrt{3}}}{3}\)
\(=\dfrac{8-\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{3\sqrt{2}}\)
\(=\dfrac{8-\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}}{3\sqrt{2}}\)
\(=\dfrac{8-\left(\sqrt{3}+1\right)+\left(\sqrt{3}-1\right)}{3\sqrt{2}}=\dfrac{6}{3\sqrt{2}}=\sqrt{2}\)
\(\sqrt{\dfrac{2+a-2\sqrt{2a}}{a+3-2\sqrt{3a}}}\)
\(=\sqrt{\dfrac{\left(\sqrt{a}-\sqrt{2}\right)^2}{\left(\sqrt{a}-\sqrt{3}\right)^2}}\)
\(=\dfrac{\left|\sqrt{a}-\sqrt{2}\right|}{\left|\sqrt{a}-\sqrt{3}\right|}\)
câu g
(câu cuối) đề nhiều trôi hết nhìn thấy mỗi câu (g)
\(G=0,1\sqrt{200}+2\sqrt{0,08}+0,4\sqrt{50}\)
\(G=0,1.10\sqrt{2}+\dfrac{2.2}{10}\sqrt{2}+0,4.5\sqrt{2}\)
\(G=\sqrt{2}\left(1+\dfrac{2}{5}+2\right)=\dfrac{\sqrt{2}\left(5+2+10\right)}{5}=\dfrac{17\sqrt{2}}{5}\)
Giải câu d thôi mấy câu còn lại đơn giản lắm nên bạn tự làm.
d/ \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
Điều kiện \(x\ge1\)
\(\Leftrightarrow\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)
\(\Leftrightarrow\sqrt{\left(2-\sqrt{x-1}\right)^2}+\sqrt{\left(3-\sqrt{x-1}\right)^2}=1\)
\(\Leftrightarrow|2-\sqrt{x-1}|+|3-\sqrt{x-1}|=1\)
Đây chỉ là phương trình cơ bản của trị tuyệt đối lớp 6, 7 học rồi nên bạn tự làm nhé.
a) \(\sqrt{2}\cdot x-\sqrt{50}=0< =>\sqrt{2}\cdot x=\sqrt{50}\)
<=> x= 5
b) \(\sqrt{3}\cdot x+\sqrt{3}=\sqrt{12}+\sqrt{27}\)
<=> \(\sqrt{3}\cdot\left(x+1\right)=\sqrt{3}\cdot\sqrt{4}+\sqrt{3}\cdot\sqrt{9}\)
<=> \(\sqrt{3}\cdot\left(x+1\right)=\sqrt{3}\cdot5< =>x+1=5\)
<=> x=4
c) \(\sqrt{3}\cdot x^2-\sqrt{12}=0\\ < =>x^2=\sqrt{4}=2;-2\\ < =>x=\sqrt{2};-\sqrt{2}\)
d) \(\dfrac{x^2}{\sqrt{5}}-\sqrt{20}=0\\ < =>x^2=\sqrt{100}=10;-10\\ < =>x=\sqrt{10};-\sqrt{10}\)