1, 3(x-2)\(\sqrt{x^2+1}\)-x2 = x-6
2, (6x2 +x+1) \(\sqrt{x-1}\)+3x2+2x-1=0
3, 2x(x-1)+1> \(\sqrt{x^2-x+1}\)
Bài 1: tìm đạo hàm của các hàm số sau
1. y=6x2 -\(\dfrac{4}{x}\)+1
2. y=\(\dfrac{2x+1}{-x+1}\)
3. y= \(\sqrt{x^2-3x+4}\)
4. y=\(\dfrac{\left(x^2-1\right)\left(x+3\right)}{x-4}\)
5. y=\(\dfrac{1}{2x^2-3x+5}\)
6. y=(x+1)\(\sqrt{x^2-1}\)
1.
\(y'=12x+\dfrac{4}{x^2}\)
2.
\(y'=\dfrac{3}{\left(-x+1\right)^2}\)
3.
\(y'=\dfrac{2x-3}{2\sqrt{x^2-3x+4}}\)
4.
\(y=\dfrac{x^3+3x^2-x-3}{x-4}\)
\(y'=\dfrac{\left(3x^2+6x-1\right)\left(x-4\right)-\left(x^3+3x^2-x-3\right)}{\left(x-4\right)^2}=\dfrac{2x^3-9x^2-24x+7}{\left(x-4\right)^2}\)
5.
\(y'=-\dfrac{4x-3}{\left(2x^2-3x+5\right)^2}\)
6.
\(y'=\sqrt{x^2-1}+\dfrac{x\left(x+1\right)}{\sqrt{x^2-1}}\)
1) (\(\dfrac{1}{2}\)x + 3)*(x2- 4x- 6)
2) (6x2 -9x +15)*(\(\dfrac{2}{3}\)x+1)
3) (3x2 -x+5)*(x3+5x-1)
4) (x-1)*(x+1)*(x-2)
5) D=(x-7)*(x+5)-(x-4)*(x+3)
6) E= 4x*(x2-x-1)-(x+3)*(x2-2)
7) F= 5x*(x-3)*(x-1)-4x*(x2-2x)
1) \(\left(\dfrac{1}{2}x+3\right)\left(x^2-4x-6\right)\)
\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)
\(=\dfrac{1}{2}x^3+x^2-15x-18\)
2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)
\(=4x^3+6x^2-6x^2-9x+10x+15\)
\(=4x^3+x+15\)
3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)
\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)
\(=3x^5-x^4+5x^3+10x^2+26x-5\)
4) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)
\(=\left(x^2-1\right)\left(x-2\right)\)
\(=x^3-2x^2-x+2\)
giải phương trình
a) \(\left(x+\frac{5-x}{\sqrt{x}+1}\right)^2+\frac{16\sqrt{x}\left(5-x\right)}{\sqrt{x}+1}-16\)\(=0\)
b) \(\sqrt{2x-\frac{3}{x}}+\sqrt{\frac{6}{x}-2x}=1+\frac{3}{2x}\)
c) \(\sqrt{2x+1}+\frac{2x-1}{x+3}-\left(2x-1\right)\sqrt{x^2+4}-\sqrt{2}=0\)
d) \(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)
1)\(\sqrt{4+2x-x^2}=x-2\)
2)\(\sqrt{25-x^2}=x-1\)
3)(x+4).\(\sqrt{10-x^2}=x^2+2x-8\)
4)(x-3).\(\sqrt{x^2-3x+2}=x^2-8x+15\)
5)\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x-6\sqrt{x-1}+8}=1\)
6)\(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)
7)\(^{x^2+x-2\sqrt{x+1}+2=0}\)
8)x-4\(\sqrt{2x+4}-2\sqrt{1-x}+10=0\)
1.
ĐK: $-x^2+2x+4\geq 0$
PT \(\Rightarrow \left\{\begin{matrix} x-2\geq 0\\ 4+2x-x^2=(x-2)^2=x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 2\\ 6x=2x^2\end{matrix}\right.\Rightarrow x=3\) (thỏa mãn)
Vậy...........
2)
ĐK: $-5\leq x\leq 5$
PT \(\Rightarrow \left\{\begin{matrix} x-1\geq 0\\ 25-x^2=(x-1)^2=x^2-2x+1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 2x^2-2x-24=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x^2-x-12=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ (x+3)(x-4)=0\end{matrix}\right.\)
\(\Rightarrow x=4\) (thỏa mãn)
3)
ĐK: $x^2\leq 10$
PT $\Leftrightarrow (x+4)\sqrt{10-x^2}=(x+4)(x-2)$
$\Leftrightarrow (x+4)[\sqrt{10-x^2}-(x-2)]=0$
Nếu $x+4=0\Rightarrow x=-4$ (không thỏa mãn ĐKXĐ)
Nếu $\sqrt{10-x^2}-(x-2)=0$
$\Leftrightarrow \sqrt{10-x^2}=x-2$
\(\Rightarrow \left\{\begin{matrix} x-2\geq 0\\ 10-x^2=(x-2)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 2\\ 2x^2-4x-6=0\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ 2(x-3)(x+1)=0\end{matrix}\right.\Rightarrow x=3\)
Giải phương trình:
1) \(x^4-2\sqrt{3}x^2+x+3-\sqrt{3}=0\)
2)\(\dfrac{1}{1+\sqrt{2x^2+1}}\)+\(\dfrac{\sqrt{x^2+1}}{1+\sqrt{x^2+1}}\)-\(\dfrac{32}{\sqrt{2\sqrt{2x^2+1}\left(1+\sqrt{2x^2+1}\right)+2\sqrt{\dfrac{1}{x^2+1}}\left(1+\sqrt{\dfrac{1}{x^2+1}}\right)+8}}\)= -7
3)\(2x^2\left(x-1\right)+x=\left(x-1\right)\sqrt{2x\left(x^2-x+2\right)}+6\)
giaỉ pt:
a, \(\sqrt{x +1}+2\left(x+1\right)=x-1+\sqrt{1-x}+3\sqrt{1-x^2}\)
b, \(14\sqrt{x+35}+6\sqrt{x+1}=84+\sqrt{x^2+36x+35}\)
c, \(x\sqrt{2x+3}+3\left(\sqrt{x+5}+1\right)=3x+\sqrt{2x^2+13x+15}+\sqrt{2x+3}\)
b.
ĐKXĐ: \(x\ge-1\)
\(\sqrt{\left(x+1\right)\left(x+35\right)}-14\sqrt{x+35}+84-6\sqrt{x+1}=0\)
\(\Leftrightarrow\sqrt{x+1}\left(\sqrt{x+35}-14\right)-6\left(\sqrt{x+35}-14\right)=0\)
\(\Leftrightarrow\left(\sqrt{x+1}-6\right)\left(\sqrt{x+35}-14\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=6\\\sqrt{x+35}=14\end{matrix}\right.\)
\(\Leftrightarrow...\)
a. ĐKXĐ: \(-1\le x\le1\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{1-x}=b\ge0\end{matrix}\right.\)
\(\Rightarrow a+2a^2=-b^2+b+3ab\)
\(\Leftrightarrow\left(2a^2-3ab+b^2\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-b\right)+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(2a-b+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\2a+1=b\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=\sqrt{1-x}\\2\sqrt{x+1}+1=\sqrt{1-x}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\4x+5+4\sqrt{x+1}=1-x\left(1\right)\end{matrix}\right.\)
(1) \(\Leftrightarrow4\sqrt{x+1}=-4-5x\) \(\left(x\le-\dfrac{4}{5}\right)\)
\(\Leftrightarrow16\left(x+1\right)=25x^2+40x+16\)
\(\Leftrightarrow25x^2+24x=0\Rightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=-\dfrac{24}{25}\end{matrix}\right.\)
c.
ĐKXĐ: \(x\ge-\dfrac{3}{2}\)
\(\Leftrightarrow x\sqrt{2x+3}-\sqrt{2x+3}+3-3x+3\sqrt{x+5}-\sqrt{\left(2x+3\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\sqrt{2x+3}\left(x-1\right)-3\left(x-1\right)-\sqrt{x+5}\left(\sqrt{2x+3}-3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\sqrt{2x+3}-3\right)-\sqrt{x+5}\left(\sqrt{2x+3}-3\right)=0\)
\(\Leftrightarrow\left(x-1-\sqrt{x+5}\right)\left(\sqrt{2x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1-\sqrt{x+5}=0\\\sqrt{2x+3}-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5-\sqrt{x+5}-6=0\\\sqrt{2x+3}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=-2\left(loại\right)\\\sqrt{x+5}=3\\\sqrt{2x+3}=3\end{matrix}\right.\)
\(\Leftrightarrow...\)
Giải các phương trình sau:
a) \(x\sqrt{x-1}+\left(2x+1\right)\sqrt{x+2}+x^3-4x^2+x-6=0\)
b) \(\left(2x+3\right)\sqrt{2x-1}+x\sqrt{x+3}+x^2-5x-3=0\)
c) \(x\sqrt{2x+3}+\left(x+1\right)\sqrt{4x-1}+2\left(x^2-x-1\right)=0\)
a) ĐKXĐ: \(x\ge1\).
\(PT\Leftrightarrow x\left(\sqrt{x-1}-1\right)+\left(2x+1\right)\left(\sqrt{x+2}-2\right)+\left(x^3-4x^2+6x-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{x}{\sqrt{x-1}+1}+\frac{2x+1}{\sqrt{x+2}+2}+x^2-2x+2\right)=0\)
\(\Leftrightarrow x=2\left(TMĐK\right)\)
b) Cách làm cũng giống như thế :v
ĐKXĐ: \(x\ge\frac{1}{2}\)
\(PT\Leftrightarrow\left(x-1\right)\left(\frac{4x+6}{\sqrt{2x-1}+1}+\frac{x}{\sqrt{x+3}+2}+x\right)=0\)
\(\Leftrightarrow x=1\) (TMĐK)
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)
giải pt
a) \(3\sqrt{x}+\frac{3}{2\sqrt{x}}=2x+\frac{1}{2x}-7\)
b) \(5\sqrt{x}+\frac{5}{2\sqrt{x}}=2x+\frac{1}{2x}+4\)
c) \(\sqrt{2x^2+8x+5}+\sqrt{2x^2-4x+5}=6\sqrt{x}\)
d) \(x+1+\sqrt{x^2-4x+1}=3\sqrt{x}\)
e) \(x^2+2x\sqrt{x-\frac{1}{x}}=3x+1\)
f) \(x^2-6x+x\sqrt{\frac{x^2-6}{x}}-6=0\)
g) \(\frac{3x^2}{3+\sqrt{x}}+6+2\sqrt{x}=5x\)
h) \(\frac{x^2}{4-3\sqrt{x}}+8=3\left(x+2\sqrt{x}\right)\)
a/ ĐKXĐ: ...
\(\Leftrightarrow3\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)-7\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow a^2=x+\frac{1}{4x}+1\)
\(\Rightarrow x+\frac{1}{4x}=a^2-1\)
Pt trở thành:
\(3a=2\left(a^2-1\right)-7\)
\(\Leftrightarrow2a^2-3a-9=9\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x}+\frac{1}{2\sqrt{x}}=3\)
\(\Leftrightarrow2x-6\sqrt{x}+1=0\)
\(\Rightarrow\sqrt{x}=\frac{3+\sqrt{7}}{2}\Rightarrow x=\frac{8+3\sqrt{7}}{2}\)
b/ ĐKXĐ:
\(\Leftrightarrow5\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)=2\left(x+\frac{1}{4x}\right)+4\)
Đặt \(\sqrt{x}+\frac{1}{2\sqrt{x}}=a>0\Rightarrow x+\frac{1}{4x}=a^2-1\)
\(\Rightarrow5a=2\left(a^2-1\right)+4\Leftrightarrow2a^2-5a+2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{1}{2}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\sqrt{x}+\frac{1}{2\sqrt{x}}=2\\\sqrt{x}+\frac{1}{2\sqrt{x}}=\frac{1}{2}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-4\sqrt{x}+1=0\\2x-\sqrt{x}+1=0\left(vn\right)\end{matrix}\right.\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)
\(\Leftrightarrow\frac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)
\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\frac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\frac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)
\(\Leftrightarrow2x^2-8x+5=0\)
d/ ĐKXĐ: ...
\(\Leftrightarrow x+1-\frac{15}{6}\sqrt{x}+\sqrt{x^2-4x+1}-\frac{1}{2}\sqrt{x}=0\)
\(\Leftrightarrow\frac{x^2-\frac{17}{4}x+1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{x^2-\frac{17}{4}x+1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}=0\)
\(\Leftrightarrow\left(x^2-\frac{17}{4}x+1\right)\left(\frac{1}{\left(x+1\right)^2+\frac{15}{6}\sqrt{x}}+\frac{1}{\sqrt{x^2-4x+1}+\frac{1}{2}\sqrt{x}}\right)=0\)
\(\Leftrightarrow x^2-\frac{17}{4}x+1=0\)
\(\Leftrightarrow4x^2-17x+4=0\)
e/ ĐKXĐ: ...
\(\Leftrightarrow x^2-1+2x\sqrt{\frac{x^2-1}{x}}=3x\)
Nhận thấy \(x=0\) không phải nghiệm, pt tương đương:
\(\frac{x^2-1}{x}+2\sqrt{\frac{x^2-1}{x}}=3\)
Đặt \(\sqrt{\frac{x^2-1}{x}}=a\ge0\)
\(a^2+2a=3\Leftrightarrow a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\frac{x^2-1}{x}}=1\Leftrightarrow x^2-1=x\Leftrightarrow x^2-x-1=0\)
f/ ĐKXĐ: ...
\(\Leftrightarrow x^2-6+x\sqrt{\frac{x^2-6}{x}}-6x=0\)
Nhận thấy \(x=0\) ko phải nghiệm, pt tương đương:
\(\frac{x^2-6}{x}+\sqrt{\frac{x^2-6}{x}}-6=0\)
Đặt \(\sqrt{\frac{x^2-6}{x}}=a\ge0\)
\(a^2+a-6=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-3\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{\frac{x^2-6}{x}}=2\Leftrightarrow x^2-4x-6=0\)
1) \(\dfrac{x-3x^2}{2}+\sqrt{2x^4-x^3+7x^2-3x+3}=2\)
2) \(1+\sqrt{\dfrac{x-2}{1-x}}=\dfrac{2x^2-2x+1}{x^2-2x+2}\)
3) \(x+y+z+\dfrac{3}{x-1}+\dfrac{3}{y-1}+\dfrac{3}{z-1}=2\left(\sqrt{x+2}+\sqrt{y+2}+\sqrt{z+2}\right)\) với x ,y ,z > 1
4) \(\sqrt[3]{x+6}+x^2=7-\sqrt{x-1}\)
5) \(x^4-2x^3+x-\sqrt{2\left(x^2-x\right)}=0\)