Tìm x :
a) 4x2 - 9 = 0
b) 2x2 + 0,36 = 1
c) \(\frac{5}{12}.\sqrt{x}-\frac{1}{6}=\frac{1}{3}\)
d) 3x2 + 7 = -4
Bài 5: Tìm nghiệm của các đa thức sau: Dạng 1: a) 4x + 9 b) -5x + 6 c) 7 – 2x d) 2x + 5 Dạng 2: a) ( x+ 5 ) ( x – 3) b) ( 2x – 6) ( x – 3) c) ( x – 2) ( 4x + 10 ) Dạng 3: a) x2 -2x b) x2 – 3x c) 3x2 – 4x d) ( 2x- 1)2 Dạng 4: a) x2 – 1 b) x2 – 9 c)– x 2 + 25 d) x2 - 2 e) 4x2 + 5 f) –x 2 – 16 g) - 4x4 – 25 Dạng 5: a) 2x2 – 5x + 3 b) 4x2 + 6x – 1 c) 2x2 + x – 1 d) 3x2 + 2x – 1
Rút gọn:
a, A = \(\frac{1}{\sqrt{3}+\sqrt{1}}+\frac{1}{\sqrt{5}+\sqrt{3}}+\frac{1}{\sqrt{7}+\sqrt{5}}+\frac{1}{\sqrt{9}+\sqrt{7}}\)
b, B = \(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
c, C = \(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}-\sqrt{6}\)
d, D = \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\) với x ≥ 2
a.
\(A=\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{3}+\sqrt{5}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{9}}\)
\(=\frac{\sqrt{3}-\sqrt{1}}{3-1}+\frac{\sqrt{5}-\sqrt{3}}{5-3}+\frac{\sqrt{7}-\sqrt{5}}{7-5}+\frac{\sqrt{9}-\sqrt{7}}{9-7}\)
\(=\frac{\sqrt{9}-\sqrt{7}+\sqrt{7}-\sqrt{5}+\sqrt{5}-\sqrt{3}+\sqrt{3}-\sqrt{1}}{2}\)
\(=\frac{3-1}{2}=1\)
b.
\(B=2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{48}}\)
\(=2\sqrt{80\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{20\sqrt{3}}\)
\(=8\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}}-6\sqrt{5\sqrt{3}}=0\)
c.
\(C=\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}-\sqrt{6}\)
\(=\frac{15\sqrt{6}-15}{6-1}+\frac{4\sqrt{6}+8}{6-4}-\frac{36+12\sqrt{6}}{9-6}-\sqrt{6}\)
\(=\frac{15\sqrt{6}-15}{5}+\frac{4\sqrt{6}+8}{2}-\frac{36+12\sqrt{6}}{3}-\sqrt{6}\)
\(=3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}-\sqrt{6}\)
\(=-11\)
d)D=\(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\)( \(x\ge2\))
=\(\sqrt{x+2\sqrt{2}.\sqrt{x-2}}+\sqrt{x-2\sqrt{2}.\sqrt{x-2}}\)
=\(\sqrt{\left(x-2\right)+2\sqrt{2}.\sqrt{x-2}+2}+\sqrt{\left(x-2\right)-2\sqrt{2}.\sqrt{x-2}+2}\)
=\(\sqrt{\left(\sqrt{x-2}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{x-2}-\sqrt{2}\right)^2}\)
=\(\sqrt{x-2}+\sqrt{2}+\left|\sqrt{x-2}-\sqrt{2}\right|\)(1)
TH1: \(2\le x\le4\)
Từ (1)<=> \(\sqrt{x-2}+\sqrt{2}-\sqrt{x-2}+\sqrt{2}\)
=\(2\sqrt{2}\)
TH2. x\(>4\)
Từ (1) <=> \(\sqrt{x-2}+\sqrt{2}-\sqrt{2}+\sqrt{x-2}\)=\(2\sqrt{x-2}\)
Vậy \(\left[{}\begin{matrix}2\le x\le4\\x>4\end{matrix}\right.< =>\left[{}\begin{matrix}D=2\sqrt{2}\\D=2\sqrt{x-2}\end{matrix}\right.\)
a) \(\frac{6^7}{4^3x9^2}\)
b) \(\frac{12^3x15^3}{4^3x25^2x9^2}\)
c) \(\frac{2^{11}+3x2^{10}}{10x4^5}\)
d) \(\frac{9^4x2-3^6}{34x3^7}\)
a) \(\frac{6^7}{4^3\cdot9^2}=\frac{2^7\cdot3^7}{2^6\cdot3^4}=2\cdot3^3=2\cdot27=54\)
b) \(\frac{12^3\cdot15^3}{4^3\cdot25^2\cdot9^2}=\frac{2^6\cdot3^3\cdot3^3\cdot5^3}{2^6\cdot5^4\cdot3^4}=\frac{3^2}{5}=1,8\)
c) \(\frac{2^{11}+3\cdot2^{10}}{10\cdot4^5}=\frac{2^{10}\left(2+3\right)}{2\cdot5\cdot2^{10}}=\frac{1}{2}=0,5\)
d) \(\frac{3^8\cdot2-3^6}{2\cdot17\cdot3^7}=\frac{3^6\left(3^2\cdot2-1\right)}{2\cdot17\cdot3^7}=\frac{1}{2\cdot3}=\frac{1}{6}\)
1. \(\sqrt{x-2\sqrt{x-1}}+\sqrt{x+3-4\sqrt{x-1}}\left(2< x< 5\right)\)
2. \(\frac{6}{1-\sqrt{3}}-\frac{3\sqrt{3}-1}{\sqrt{3}+1}+\sqrt{3}\)
3. \(\sqrt{29-12\sqrt{5}+\sqrt{24-8\sqrt{3}}}\)
4. \(\sqrt{\frac{4}{9-4\sqrt{5}}}-\sqrt{\frac{4}{9+4\sqrt{5}}}\)
5. \(5\sqrt{\frac{1}{5}}+\frac{1}{2}\sqrt{x}-\frac{5}{4}\sqrt{\frac{4}{5}+\sqrt{5}}\)
6. \(\frac{6-\sqrt{6}}{\sqrt{6}-1}-9\sqrt{\frac{2}{3}}-\frac{4}{2-\sqrt{6}}\)
7. \(\left(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\frac{\left(\sqrt{x}-1\right)^2}{2}\left(x\ge0,x\ne1\right)\)
Trả lời nhanh giúp mình với mình cần gấp lắm
tìm x biết
a)\(\frac{3\sqrt{x}-5}{2}-\frac{2\sqrt{x}-7}{3}+1=\sqrt{x}\)
b)\(\sqrt{9x^2+45}-\frac{1}{12}\sqrt{16x^2+80}+3\sqrt{\frac{x^2+5}{16}}-\frac{1}{4}\sqrt{\frac{25x^2+125}{9}}=9\)
bài 1) rút gọn
1) 5√\(\frac{1}{5}\) 2)\(\frac{12}{5}\)√\(\frac{5}{4}\) 3)\(\frac{30}{5\sqrt{6}}\) 4) \(\frac{20}{2\sqrt{5}}\) 5)\(\frac{2-\sqrt{2}}{\sqrt{2}}\) 6) \(\frac{11+\sqrt{11}}{1+\sqrt{ }11}\) 7) \(\frac{\sqrt{21-\sqrt{7}}}{1-\sqrt{3}}\) 8)\(\frac{\sqrt{2+\sqrt{3}}}{2+\sqrt{6}}\) 9)\(\frac{\sqrt{10-\sqrt{2}}}{\sqrt{5-}1}\) 10)\(\frac{2\sqrt{3}-3\sqrt{2}}{\sqrt{3}-\sqrt[]{2}}\)
bài 2) với các biểu thức đã cho là có nghĩa và rút gọn
1)\(\frac{x-\sqrt{x}}{\sqrt{x}-1}\) 2)\(\frac{x\sqrt{x}-2x}{2-\sqrt{x}}\) 3) \(\frac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\) 4) \(\frac{a\sqrt{b}-\sqrt{a}}{\sqrt{b}-b\sqrt{a}}\) 5) \(\frac{a-1}{\sqrt{a}+1}\) 6) \(\frac{4-x}{2\sqrt{x}-x}\) 7)\(\frac{a+1+2\sqrt{a}}{1+\sqrt{a}}\) 8)\(\frac{3\sqrt{x}-x}{3+2\sqrt{3x}-x}\) 9)\(\frac{y+12-4\sqrt{3y}}{y-12}\) 10)\(\frac{4\sqrt{x}-x-4}{x-4}\) 11)\(\frac{x+y-2\sqrt{xy}}{x\sqrt{y}-y\sqrt{x}}\)
Giúp vs ạ
Bài 1 giải các bất phương trình sau
a.x2 - x - 6 = 0
b.2x2 - 7x + 5 < 0
c.3x2 - 9x + 6 ≥ 0
d.2x2 - 5x + 3 < 0
Bài 2 Giải phương trình sau
A.√x2 + x + 5 = √2x2 - 4x + 1
B.√11x2 -14x - 12 = √3x2 + 4x - 7
Bài 2:
a: =>2x^2-4x+1=x^2+x+5
=>x^2-5x-4=0
=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)
b: =>11x^2-14x-12=3x^2+4x-7
=>8x^2-18x-5=0
=>x=5/2 hoặc x=-1/4
bài 15 tìm x biết
a\(\frac{x}{4}-\frac{3}{7}+\frac{2}{5}=\frac{31}{140}\)
b\(\frac{5}{12}+\frac{5}{x}-\frac{1}{8}=\frac{1}{2}\)
c\(x+\frac{1}{9}-\frac{3}{5}=\frac{3}{6}\)
d\(\frac{3}{4}-x+\frac{6}{-11}=\frac{5}{6}\)
e\(x-\frac{1}{9}-\frac{3}{5}=\frac{3}{6}\)
gpt:a, \(x+\)\(\frac{4x}{x+4\sqrt{x}+4}\)\(=12\)
b,\(x+\frac{x}{\sqrt{x^2-1}}=\frac{35}{12}\)
c,\(x^2+\sqrt{x+7}=7\)
d,\(\sqrt{\frac{x-6}{3}}+\sqrt{\frac{x-5}{4}}+\sqrt{\frac{x-7}{2}}=\sqrt{\frac{x-3}{6}}+\sqrt{\frac{x-4}{5}}+\sqrt{\frac{x-2}{7}}\)
e,cho a,b >0 và \(a^3+b^3+6ab\le8\).Tìm GTNN của A=\(\frac{1}{a}+\frac{1}{b}\)
f,cho a>0 .Tìm GTNN của B=\(9a+\frac{1}{9a}-\frac{6\sqrt{a}+8}{a+1}+2020\)
g,cho hình vuông ABCD có AC giao BD tại E ,\(I\in AB,M\in BC\)sao cho góc IEM =90 ,AM giao DC tại N, BN giao EM tại K .CM:\(CK\perp BN\)