\(\sqrt[4]{56-x}+\sqrt[4]{x+41}=5\)
tìm x thoả mãn: \(x^3+\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}x+5=0\)
Xét \(\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45-4\sqrt{41}}}=\frac{8\sqrt{41}}{\sqrt{\left(\sqrt{41}+2\right)^2}+\sqrt{\left(\sqrt{41}-2\right)^2}}=\frac{8\sqrt{41}}{\sqrt{41}+2+\sqrt{41}-2}=\frac{8\sqrt{41}}{2\sqrt{41}}=4\)
Phương trình trên tương đương:
x3+4x+5=0
<=>x(x2-1)+5(x+1)=0
<=>x(x-1)(x+1)+5(x+1)=0
<=>(x+1)(x2-x+5)=0
<=>x+1=0 hoặc x2-x+5=0(vô nghiệm)
<=>x=-1
Vậy pt trên có nghiệm là x=-1
Bài này đi thi vio mk cũng gặp ..
bằng 1 ak
Mọi người giúp mình mấy câu rút gọn này với
A= \(\frac{8\sqrt{41}}{\sqrt{45+4\sqrt{41}}+\sqrt{45+4\sqrt{41}}}\)
B= \(\left(\frac{2x+1}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right)\div\left(1-\frac{x+4}{x+\sqrt{x}+1}\right)\)
A = \(\frac{8\sqrt{41}}{2\sqrt{2^2+2.2.\sqrt{41}+\sqrt{41}^2}}\)
A = \(\frac{8\sqrt{41}}{2\sqrt{\left(2+\sqrt{41}\right)^2}}\)
A = \(\frac{8\sqrt{41}}{2\left|2+\sqrt{41}\right|}\)
A = \(\frac{8\sqrt{41}}{4+2\sqrt{41}}\)
B = \(\left(\frac{2x+1}{\sqrt{x}^3+1^3}-\frac{1}{\sqrt{x}-1}\right):\frac{x+\sqrt{x}+1+x+4}{x+\sqrt{x}+1}\)
B = \(\left(\frac{2x+1}{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}-1}\right).\frac{x+\sqrt{x}+1}{2x+\sqrt{x}+5}\)
Bạn tự làm tiếp nhé, mỏi tay quá!!
\(A=\frac{8\sqrt{41}}{2\sqrt{45+4\sqrt{41}}}=\frac{8\sqrt{41}}{2\sqrt{41+4\sqrt{41}+4}}=\frac{8\sqrt{41}}{2\sqrt{\left(\sqrt{41}\right)^2+2\cdot\sqrt{41}\cdot2+2^2}}\)
\(=\frac{8\sqrt{41}}{2\sqrt{\left(\sqrt{41}+2\right)^2}}=\frac{8\sqrt{41}}{2\left(\sqrt{41}+2\right)}=\frac{8\sqrt{41}\left(\sqrt{41}-2\right)}{2\left(41-4\right)}=\frac{328-16\sqrt{41}}{74}=\frac{164-8\sqrt{41}}{37}\)
\(B=\left(\frac{2x+1}{x\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right):\left(1-\frac{x+4}{x+\sqrt{x}+1}\right)\)
\(=\left(\frac{2x+1}{\sqrt{x}^3+1^3}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{x+\sqrt{x}+1-x-4}{x+\sqrt{x}+1}\right)\)
\(=\left(\frac{2x+1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right):\left(\frac{\sqrt{x}-3}{x+\sqrt{x}+1}\right)\)
\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\frac{x+\sqrt{x}+1}{\sqrt{x}-3}\)
\(=\frac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)}=\frac{\sqrt{x}}{\sqrt{x}-3}=\frac{\sqrt{x}\left(\sqrt{x}+3\right)}{x-9}=\frac{x+3\sqrt{x}}{x-9}\)
rút gọn
1, \(\sqrt{8-\sqrt{35}}\)
2, \(4\sqrt{3+2\sqrt{2}}-\sqrt{56\sqrt{2}+81}\)
3, \(\frac{x-49}{\sqrt{x}-7}\)
4, \(\sqrt{x+2\sqrt{x+1}}\)
5, \(\sqrt{x-1-2\sqrt{x-2}}\)
2.Ta có : \(4\sqrt{3+2\sqrt{2}}-\sqrt{56\sqrt{2}+81}\)
\(=4\sqrt{2+2\sqrt{2}+1}-\sqrt{56\sqrt{2}+81}\)
\(=4\sqrt{2}+4-\sqrt{56\sqrt{2}+81}\)
\(=4\sqrt{2}+4-\sqrt{7^2+2.4\sqrt{2}.7+\left(4\sqrt{2}\right)^2}\)
\(=4\sqrt{2}+4-7-4\sqrt{2}=4-7=-3\)
3.Ta có : \(\frac{x-49}{\sqrt{x}-7}\)
\(=\frac{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}{\sqrt{x}-7}=\sqrt{x}+7\)
4.Ta có : \(\sqrt{x+2\sqrt{x+1}}\)
\(=\sqrt{x+1+2\sqrt{x+1}+1-1}\)
\(=\sqrt{\left(\sqrt{x+1}+1\right)^2-1}\)
5.Ta có : \(\sqrt{x-1-2\sqrt{x-2}}\)
\(=\sqrt{x-2-2\sqrt{x-2}+1}\)
\(=\sqrt{\left(\sqrt{x-2}-1\right)^2}=\left|\sqrt{x-2}-1\right|\)
Tính giá trị
B= \(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
C=\(\sqrt{21+\sqrt{41}}.\sqrt{5+\sqrt{4+\sqrt{41}}}.\sqrt{3+\sqrt{4+\sqrt{4+\sqrt{41}}}}.\sqrt{3-\sqrt{4+\sqrt{4+\sqrt{41}}}}\)
B= \(\sqrt{2+\sqrt{3}}.\sqrt{2+\sqrt{2+\sqrt{3}}}.\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{3}}}}.\sqrt{2-\sqrt{2+\sqrt{2+\sqrt{3}}}}\)
C=\(\sqrt{21+\sqrt{41}}.\sqrt{5+\sqrt{4+\sqrt{41}}}.\sqrt{3+\sqrt{4+\sqrt{4+\sqrt{41}}}}.\sqrt{3-\sqrt{4+\sqrt{4+\sqrt{41}}}}\)
Diễn giải cho t vs nhé :) camon's
B=1 :') ; C =23.22760565 ?
Btw : Tất cả đều nhờ máy tính =))
Cho ΔABC vuông tại A có AB=5;AC=4.Bán kính đường tròn qua A và tiếp xúc với BC tại B bằng
A.\(\dfrac{5}{4}\sqrt{41}\) B.\(\dfrac{5}{2}\sqrt{41}\) C.\(\sqrt{41}\) D.\(\dfrac{5}{8}\sqrt{41}\)
Tìm x
a)\(4\sqrt{9x-63}+\dfrac{\sqrt{8x-56}}{\sqrt{2}}=42\)
b)\(\sqrt{x^2-2\sqrt{3x}+3}=\sqrt{10+2\sqrt{21}}\)
c)\(x\sqrt{x-4}-3\sqrt{25x-100}=0\)
a: \(\Leftrightarrow12\sqrt{x-7}+\sqrt{4x-28}=42\)
\(\Leftrightarrow14\sqrt{x-7}=42\)
=>x-7=9
hay x=16
c: \(\Leftrightarrow x\sqrt{x-4}-15\sqrt{x-4}=0\)
=>x=4 hoặc x=15
1, \(\sqrt{x-1}+\sqrt{x-4}=5\)
2, \(2x-7\sqrt{x}+5=0\)
3, \(\sqrt{2x+1}+\sqrt{x-3}=2\sqrt{x}\)
4, \(x-4\sqrt{x}+2021\sqrt{x-4}+4=0\)
5, \(\sqrt{2x-3}-\sqrt{x+1}=7\left(4-x\right)\)
1. ĐKXĐ: $x\geq 4$
PT $\Leftrightarrow \sqrt{x-1}=5-\sqrt{x-4}$
$\Rightarrow x-1=25+x-4-10\sqrt{x-4}$
$\Leftrightarrow 22=10\sqrt{x-4}$
$\Leftrightarrow 2,2=\sqrt{x-4}$
$\Leftrightarrow 4,84=x-4\Leftrightarrow x=8,84$
(thỏa mãn)
2. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow (2x-2\sqrt{x})-(5\sqrt{x}-5)=0$
$\Leftrightarrow 2\sqrt{x}(\sqrt{x}-1)-5(\sqrt{x}-1)=0$
$\Leftrightarrow (\sqrt{x}-1)(2\sqrt{x}-5)=0$
$\Leftrightarrow \sqrt{x}-1=0$ hoặc $2\sqrt{x}-5=0$
$\Leftrightarrow x=1$ hoặc $x=\frac{25}{4}$ (tm)
3. ĐKXĐ: $x\geq 3$
Bình phương 2 vế thu được:
$3x-2+2\sqrt{(2x+1)(x-3)}=4x$
$\Leftrightarrow 2\sqrt{(2x+1)(x-3)}=x+2$
$\Leftrightarrow 4(2x+1)(x-3)=(x+2)^2$
$\Leftrightarrow 4(2x^2-5x-3)=x^2+4x+4$
$\Leftrightarrow 7x^2-24x-16=0$
$\Leftrightarrow (x-4)(7x+4)=0$
Do $x\geq 3$ nên $x=4$
Thử lại thấy thỏa mãn
Vậy $x=4$
4. ĐKXĐ: $x\geq 4$
PT $\Leftrightarrow (x-4\sqrt{x}+4)+2021\sqrt{x-4}=0$
$\Leftrightarrow (\sqrt{x}-2)^2+2021\sqrt{x-4}=0$
Ta thấy, với mọi $x\geq 4$ thì:
$(\sqrt{x}-2)^2\ge 0$
$2021\sqrt{x-4}\geq 0$
Do đó để tổng của chúng bằng $0$ thì:
$\sqrt{x}-2=\sqrt{x-4}=0$
$\Leftrightarrow x=4$ (tm)
1) \(4\left(x+1\right)^2=2.\sqrt{x^4+x^2+1}\)
2) \(\sqrt{\frac{x-56}{16}}+\sqrt{x-3}=\frac{x}{8}\)