Giai phương trình:
\(\sqrt[3]{x^2}-2\sqrt[3]{x}-\left(x-4\right)\sqrt{x-7}-3x+28=0\)
Giải các phương trình sau:
a \(x^2-11=0\)
b \(x^2-12x+52=0\)
c \(x^2-3x-28=0\)
d \(x^2-11x+38=0\)
e \(6x^2+71x+175=0\)
f \(x^2-\left(\sqrt{2}+\sqrt{8}\right)x+4=0\)
g\(\left(1+\sqrt{3}\right)x^2-\left(2\sqrt{3}+1\right)x+\sqrt{3}=0\)
a.
$x^2-11=0$
$\Leftrightarrow x^2=11$
$\Leftrightarrow x=\pm \sqrt{11}$
b. $x^2-12x+52=0$
$\Leftrightarrow (x^2-12x+36)+16=0$
$\Leftrightarrow (x-6)^2=-16< 0$ (vô lý)
Vậy pt vô nghiệm.
c.
$x^2-3x-28=0$
$\Leftrightarrow x^2+4x-7x-28=0$
$\Leftrightarrow x(x+4)-7(x+4)=0$
$\Leftrightarrow (x+4)(x-7)=0$
$\Leftrightarrow x+4=0$ hoặc $x-7=0$
$\Leftrightarrow x=-4$ hoặc $x=7$
d.
$x^2-11x+38=0$
$\Leftrightarrow (x^2-11x+5,5^2)+7,75=0$
$\Leftrightarrow (x-5,5)^2=-7,75< 0$ (vô lý)
Vậy pt vô nghiệm
e.
$6x^2+71x+175=0$
$\Leftrightarrow 6x^2+21x+50x+175=0$
$\Leftrightarrow 3x(2x+7)+25(2x+7)=0$
$\Leftrightarrow (3x+25)(2x+7)=0$
$\Leftrightarrow 3x+25=0$ hoặc $2x+7=0$
$\Leftrightarrow x=-\frac{25}{3}$ hoặc $x=-\frac{7}{2}$
f.
$x^2-(\sqrt{2}+\sqrt{8})x+4=0$
$\Leftrightarrow x^2-\sqrt{2}x-2\sqrt{2}x+4=0$
$\Leftrightarrow x(x-\sqrt{2})-2\sqrt{2}(x-\sqrt{2})=0$
$\Leftrightarrow (x-\sqrt{2})(x-2\sqrt{2})=0$
$\Leftrightarrow x-\sqrt{2}=0$ hoặc $x-2\sqrt{2}=0$
$\Leftrightarrow x=\sqrt{2}$ hoặc $x=2\sqrt{2}$
g.
$(1+\sqrt{3})x^2-(2\sqrt{3}+1)x+\sqrt{3}=0$
$\Leftrightarrow (1+\sqrt{3})x^2-(1+\sqrt{3})x-(\sqrt{3}x-\sqrt{3})=0$
$\Leftrightarrow (1+\sqrt{3})x(x-1)-\sqrt{3}(x-1)=0$
$\Leftrightarrow (x-1)[(1+\sqrt{3})x-\sqrt{3}]=0$
$\Leftrightarrow x-1=0$ hoặc $(1+\sqrt{3})x-\sqrt{3}=0$
$\Leftrightarrow x=1$ hoặc $x=\frac{3-\sqrt{3}}{2}$
Giải các phương trình sau:
1) \(\sqrt{3x^2+5x+8}-\sqrt{3x^2+5x+1}=1\)
2) \(x^2-2x-12+4\sqrt{\left(4-x\right)\left(2+x\right)}=0\)
3) \(3\sqrt{x}+\dfrac{3}{2\sqrt{x}}=2x+\dfrac{1}{2x}-7\)
4) \(\sqrt{x}-\dfrac{4}{\sqrt{x+2}}+\sqrt{x+2}=0\)
5)\(\left(x-7\right)\sqrt{\dfrac{x+3}{x-7}}=x+4\)
6) \(2\sqrt{x-4}+\sqrt{x-1}=\sqrt{2x-3}+\sqrt{4x-16}\)
7) \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\dfrac{x+3}{2}\)
Giúp mình với ajk, mink đang cần gấp
Giải phương trình:
\(28+\sqrt[3]{x^2}=3x+2\sqrt[3]{x}+\left(x-4\right)\sqrt{x-7}\)
Giải phương trình:
\(a,\sqrt{5x^3-1}+\sqrt[3]{2x-1}+x-4=0\)
\(b,\sqrt{x-2}+\sqrt{4-x}=2x^2-5x-3\)
\(c,\sqrt[3]{x^2}-2\sqrt[3]{x}-\left(x-4\right)\sqrt{x-7}-3x+28=0\)
Giai các bất phương trình sau đây :
a/ \(\sqrt{\left(x-3\right)\left(8-x\right)}+26>-x^2+11x\)
b/ \(\left(x+5\right)\left(x-2\right)+3\sqrt{x\left(x+3\right)}>0\)
c/ \(\left(x+1\right)\left(x+4\right)< 5\sqrt{x^2+5x+28}\)
d/ \(\sqrt{3x^2+5x+7}-\sqrt{3x^2+5x+2}\ge1\)
HELP ME !!!!!!
Giai phuong trinh:
\(28+\sqrt[3]{x^2}=3x+2\sqrt[3]{x}+\left(x-4\right)\sqrt{x-7}\)
Giải phương trình
\(-3x^2+x+3+\left(\sqrt{3x+2}-4\right)\sqrt{3x-2x^2}+\left(x-1\right)\sqrt{3x+2}=0\)
giải phương trình :
a, \(\left(x+1\right)\sqrt{x+8}=x^2+x+4\)
b, \(\left(2x+7\right)\sqrt{2x+7}=x^2+9x+7\)
c, \(\left(3x+1\right)\sqrt{x^2+3}=3x^2+2x+3\)
c.
\(\Leftrightarrow x^2+3-\left(3x+1\right)\sqrt{x^2+3}+2x^2+2x=0\)
Đặt \(\sqrt{x^2+3}=t>0\)
\(\Rightarrow t^2-\left(3x+1\right)t+2x^2+2x=0\)
\(\Delta=\left(3x+1\right)^2-4\left(2x^2+2x\right)=\left(x-1\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{3x+1-x+1}{2}=x+1\\t=\dfrac{3x+1+x-1}{2}=2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+3}=x+1\left(x\ge-1\right)\\\sqrt{x^2+3}=2x\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3=x^2+2x+1\left(x\ge-1\right)\\x^2+3=4x^2\left(x\ge0\right)\end{matrix}\right.\)
\(\Leftrightarrow x=1\)
a.
Đề bài ko chính xác, pt này ko giải được
b.
ĐKXĐ: \(x\ge-\dfrac{7}{2}\)
\(2x+7-\left(2x+7\right)\sqrt{2x+7}+x^2+7x=0\)
Đặt \(\sqrt{2x+7}=t\ge0\)
\(\Rightarrow t^2-\left(2x+7\right)t+x^2+7x=0\)
\(\Delta=\left(2x+7\right)^2-4\left(x^2+7x\right)=49\)
\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{2x+7-7}{2}=x\\t=\dfrac{2x+7+7}{2}=x+7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x+7}=x\left(x\ge0\right)\\\sqrt{2x+7}=x+7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-7=0\left(x\ge0\right)\\x^2+12x+42=0\left(vn\right)\end{matrix}\right.\)
\(\Rightarrow x=1+2\sqrt{2}\)
giải phương trình sau:
a) \(4x^2+\left(8x-4\right).\sqrt{x}-1=3x+2\sqrt{2x^2+5x-3}\)
b) \(8x^3-36x^2+\left(1-3x\right)\sqrt{3x-2}-3\sqrt{3x-2}+63x-32=0\)
c) \(2\sqrt[3]{3x-2}-3\sqrt{6-5x}+16=0\)
d) \(\sqrt[3]{x+6}-2\sqrt{x-1}=4-x^2\)