\(\frac{\sqrt{b^{^2}-4b+4}}{b-2}\)

b/    |x+4| - \(...">

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14 tháng 7 2017

\(a,\sqrt{1-4a+4a^2}-2a\)

\(=\sqrt{\left(1-2a\right)^2}-2a\)

\(=1-2a-2a\)

\(=1-4a\)

\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)

\(=x-2y-\sqrt{\left(x-2y\right)^2}\)

\(=x-2y-\left(x-2y\right)\)

\(=x-2y-x+2y\)

\(=0\)

\(c,x^2+\sqrt{x^4-8x^2+16}\)

\(=x^2+\sqrt{\left(x^2-4\right)^2}\)

\(=x^2+x^2-4\)

\(=2x^2-4\)

Các câu còn lại tương tự nha

14 tháng 7 2017

\(a,\sqrt{1-4a+4a^2}-2a\)

\(=\sqrt{\left(1-2a\right)^2}-2a\)

\(=\left(1-2a\right)-2a\)

\(=1-4a\)

\(b,x-2y-\sqrt{x^2-4xy+4y^2}\)

\(=x-2y-\sqrt{\left(x-2y\right)^2}\)

\(=x-2y-\left(x-2y\right)\)

\(=x-2y-x+2y\)

\(=0\)

\(c,x^2+\sqrt{x^4-8x^2+16}\)

\(=x^2+\sqrt{\left(x^2-2^2\right)^2}\)

\(=x^2+\left(x^2-4\right)\)

\(=x^2+x^2-4\)

\(=2x^2-4\)

\(d,2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}\)

\(=2x-1-\frac{\sqrt{\left(x-5\right)^2}}{x-5}\)

\(=2x-1-\frac{x-5}{x-5}\)

\(=2x-1-1\)

\(=2x-2\)

\(=2\left(x-1\right)\)

Bài 1: Tính a) \(\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}\) b) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\) c) \(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\) d) \(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\) e) \(\sqrt{\frac{8+\sqrt{15}}{2}}+\sqrt{\frac{8-\sqrt{15}}{2}}\) Bài 2: Giải pt: a) \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\) b) \(\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\) c)...
Đọc tiếp

Bài 1: Tính

a) \(\sqrt{9-\sqrt{17}}\cdot\sqrt{9+\sqrt{17}}\)

b) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)

c) \(\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2\)

d) \(\sqrt{2+\sqrt{3}}\cdot\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}\cdot\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)

e) \(\sqrt{\frac{8+\sqrt{15}}{2}}+\sqrt{\frac{8-\sqrt{15}}{2}}\)

Bài 2: Giải pt:

a) \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)

b) \(\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\)

c) \(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)

d) \(\sqrt{x-4}+\sqrt{6-x}=x^2-10x+27\)

e) \(\sqrt{2x+1}+\sqrt{17-2x}=x^4-8x^3+17x^2-8x+22\)

f) \(\sqrt{x+x^2}+\sqrt{x-x^2}=x+1\)

g) \(\sqrt{3x^2+12x+16}+\sqrt{y^2-4y+13}=5\)

Bài 3: Cho biểu thức:

P= \(\frac{\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}}{\sqrt{\frac{16}{x^2}-\frac{8}{x}+1}}\)

a) Rút gon P

b) Tìm x để P đạt GTNN, tìm GTNN đó.

c) Tìm x \(\in\) Z để P \(\in\) Z

@Nguyễn Văn Đạt@Akai Haruma Help me please~~~~ Giải thích cẩn thân hộ với.

3
21 tháng 7 2019
https://i.imgur.com/FpJWAoR.jpg

Tag nhầm người rồi anh ơi !! Em mới lớp 7 không biết mấy cái này

4 tháng 7 2017

b)\(\frac{4}{x}+\sqrt{x-\frac{1}{x}}=x+\sqrt{2x-\frac{5}{x}}\)

\(pt\Leftrightarrow\frac{4}{x}+\sqrt{x-\frac{1}{x}}-\sqrt{\frac{3}{2}}=x+\sqrt{2x-\frac{5}{x}}-\sqrt{\frac{3}{2}}\)

\(\Leftrightarrow\left(\frac{4}{x}-x\right)+\frac{x-\frac{1}{x}-\frac{3}{2}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}=\frac{2x-\frac{5}{x}-\frac{3}{2}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}\)

\(\Leftrightarrow\frac{-\left(x-2\right)\left(x+2\right)}{x}+\frac{\frac{\left(x-2\right)\left(2x+1\right)}{2x}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}-\frac{\frac{\left(x-2\right)\left(4x+5\right)}{2x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}=0\)

\(\Leftrightarrow\left(x-2\right)\left(\frac{-\left(x+2\right)}{x}+\frac{\frac{\left(2x+1\right)}{2x}}{\sqrt{x-\frac{1}{x}}+\sqrt{\frac{3}{2}}}-\frac{\frac{\left(4x+5\right)}{2x}}{\sqrt{2x-\frac{5}{x}}+\sqrt{\frac{3}{2}}}\right)=0\)

Pt trong ngoặc VN suy ra x=2

4 tháng 7 2017

a)\(x^2+3\sqrt{x^2-1}=\sqrt{x^4-x^2+1}\)

\(\Leftrightarrow x^2+3\sqrt{x^2-1}-1=\sqrt{x^4-x^2+1}-1\)

\(\Leftrightarrow\frac{x^2\left(3\sqrt{x^2-1}+1\right)}{3\sqrt{x^2-1}+1}+\frac{9\left(x^2-1\right)-1}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2+1-1}{\sqrt{x^4-x^2+1}+1}\)

\(\Leftrightarrow\frac{9x^2-10+3x^2\sqrt{x^2-1}+x^2}{3\sqrt{x^2-1}+1}=\frac{x^4-x^2}{\sqrt{x^4-x^2+1}+1}\)

\(\Leftrightarrow\frac{\sqrt{x^2-1}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}=\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}\)

\(\Leftrightarrow\frac{\sqrt{\left(x-1\right)\left(x+1\right)}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2\left(x-1\right)\left(x+1\right)}{\sqrt{x^4-x^2+1}+1}=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(\frac{\frac{1}{\sqrt{x^2-1}}\left(3x^2+10\sqrt{x^2-1}\right)}{3\sqrt{x^2-1}+1}-\frac{x^2}{\sqrt{x^4-x^2+1}+1}\right)=0\)

pt trong căn vô nghiệm

suy ra x=1; x=-1

13 tháng 3 2020

\(B=\frac{x}{x-16}+\frac{2}{\sqrt{x}-4}+\frac{2}{\sqrt{x}+4}\)

\(=\frac{x}{x-16}+\frac{2\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}+\frac{2\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)}\)

\(=\frac{x}{x-16}+\frac{2\sqrt{x}+8}{x-16}+\frac{2\sqrt{x}-8}{x-16}\)

\(=\frac{x+4\sqrt{x}}{x-16}=\frac{\sqrt{x}\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+4\right)}=\frac{\sqrt{x}}{\sqrt{x}-4}\)

\(A=2\sqrt{12}-\sqrt{75}+\sqrt{\left(\sqrt{3}-2\right)^2}\)

\(=2\sqrt{12}-\sqrt{75}+\left(2-\sqrt{3}\right)\)(vì \(\sqrt{3}< \sqrt{4}=2\))

\(\Rightarrow\frac{1}{2}A=\sqrt{12}-\frac{\sqrt{75}}{2}+1-\frac{\sqrt{3}}{2}\)

\(=\sqrt{12}+1-\frac{\sqrt{3}\left(1+5\right)}{2}=\sqrt{12}-3\sqrt{3}+1\)

\(=\sqrt{3}+1\)

\(B-\frac{1}{2}A=0\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-4}=\sqrt{3}+1\)

\(\Leftrightarrow\sqrt{x}=\left(\sqrt{3}+1\right)\left(\sqrt{x}-4\right)\)

\(\Leftrightarrow\sqrt{x}=\sqrt{3x}+\sqrt{x}-4\sqrt{x}-4\)

\(\Leftrightarrow\sqrt{3x}-4\sqrt{x}-4=0\)

\(\Leftrightarrow\sqrt{x}\left(\sqrt{3}-4\right)=4\Leftrightarrow\sqrt{x}=\frac{4}{\sqrt{3}-4}\)

\(\Rightarrow x=\left(\frac{4}{\sqrt{3}-4}\right)^2=\frac{304+128\sqrt{3}}{-173}\)

13 tháng 3 2020

Mù mịt quá, sửa từ dòng 7 từ dưới lên 

\(=-\sqrt{3}+1\)

\(B-\frac{1}{2}A=0\Leftrightarrow\frac{\sqrt{x}}{\sqrt{x}-4}=-\sqrt{3}+1\)

\(\Leftrightarrow\sqrt{x}=\left(\sqrt{x}-4\right)\left(1-\sqrt{3}\right)\)

\(\Leftrightarrow\sqrt{x}=\sqrt{x}-4-\sqrt{3x}+4\sqrt{3}\)

\(\Leftrightarrow-4-\sqrt{3x}+4\sqrt{3}=0\)

\(\Leftrightarrow\sqrt{3x}=4\sqrt{3}-4\)

\(\Leftrightarrow\sqrt{x}=\frac{4\left(\sqrt{3}-1\right)}{\sqrt{3}}\)

\(\Leftrightarrow x=\frac{64-32\sqrt{3}}{3}\)