Gỉai phương trình
\(\sqrt[3]{x^2-1}+\sqrt{3x^3-2}=3x-2\)
Gỉai các phương trình:
a) \(\sqrt{1-6X+9X^2}\) = 9
b) \(\sqrt{2X-3}\) - \(\sqrt{x+1}\) = 0
c) \(\sqrt{9x^2+12x+4}\) - 2= 3x
a) \(\sqrt{1-6x+9x^2}=9\)
\(\Leftrightarrow\sqrt{\left(1-3x\right)^2}=9\)
\(\Leftrightarrow\left|1-3x\right|=9\)
\(\Leftrightarrow\left[{}\begin{matrix}1-3x=9\\1-3x=-9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=1-9\\3x=1+9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=-8\\3x=10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{8}{3}\\x=\dfrac{10}{3}\end{matrix}\right.\)
b) \(\sqrt{2x-3}-\sqrt{x+1}=0\) (\(x\ge\dfrac{3}{2}\))
\(\Leftrightarrow\sqrt{2x-3}=\sqrt{x+1}\)
\(\Leftrightarrow2x-3=x+1\)
\(\Leftrightarrow2x-x=1+3\)
\(\Leftrightarrow x=4\left(tm\right)\)
c) \(\sqrt{9x^2+12+4}-2=3x\)
\(\Leftrightarrow\sqrt{\left(3x+2\right)^2}=3x+2\)
\(\Leftrightarrow\left|3x+2\right|=3x+2\)
\(\Leftrightarrow3x+2\ge0\)
\(\Leftrightarrow3x\ge-2\)
\(\Leftrightarrow x\ge-\dfrac{2}{3}\)
a: =>|3x-1|=9
=>3x-1=9 hoặc 3x-1=-9
=>x=-8/3 hoặc x=10/3
b: =>căn 2x-3=căn x+1
=>2x-3=x+1
=>x=4
c: =>|3x+2|=3x+2
=>3x+2>=0
=>x>=-2/3
a) Gỉai phương trình sau :
\(3x^2-2x\sqrt{3}-3=0\)
b) Gỉai hệ phương trình sau :
\(\hept{\begin{cases}x\left(x-1\right)+y=\left(x+1\right)\left(x-3\right)\\2x-3y=-1\end{cases}}\)
Gỉai Phương trình:
a) \(3x-1-\sqrt{4x^2-12x+9}\)
b) \(\sqrt{3-2\sqrt{2}}-\sqrt{x^2-2x\sqrt{3}+3}\)
bn nào trả mình VP để mình làm với TT
Gỉai Phương trình:
a) \(3x-1-\sqrt{4x^2-12x+9}=0\)
b) \(\sqrt{3-2\sqrt{2}}-\sqrt{x^2-2x\sqrt{3+3}}=0\)
a) \(3x-1-\sqrt{4x^2-12x+9}=0\)
\(\Leftrightarrow3x-1=\sqrt{4x^2-12x+9}\)
\(\Leftrightarrow3x-1=\sqrt{\left(2x-3\right)^2}=2x-3\)
\(\Leftrightarrow3x-2x=-3+1\)
\(\Leftrightarrow x=-2\)
b) Đề đúng:
\(\sqrt{3-2\sqrt{2}}-\sqrt{x^2-2x\sqrt{3}+3}=0\)
\(\Leftrightarrow\sqrt{3-2\sqrt{2}}=\sqrt{x^2-2x\sqrt{3}+3}\)
\(\Leftrightarrow\sqrt{3-2\sqrt{2}}=\sqrt{\left(x-\sqrt{3}\right)^2}=x-\sqrt{3}\)
\(\Leftrightarrow3-2\sqrt{2}=x^2-2\sqrt{3}\cdot x+3\)
\(\Leftrightarrow-x^2+2\sqrt{3}\cdot x-2\sqrt{2}=0\)
Giải pt bậc 2 có:
\(\Delta=\left(2\sqrt{3}\right)^2-4\cdot\left(-1\right)\cdot\left(-2\sqrt{2}\right)=12-8\sqrt{2}\)
=> \(\left\{{}\begin{matrix}x_1=-\dfrac{-2\sqrt{3}+\sqrt{12-8\sqrt{2}}}{2}\\x_2=-\dfrac{-2\sqrt{3}-\sqrt{12-8\sqrt{2}}}{2}\end{matrix}\right.\)
Vậy...........................
Giải bất phương trình:
\(\sqrt{3x^2-7x+3}+\sqrt{x^2-3x+4}>\sqrt{x^2-2}+\sqrt{3x^2-5x-1}\)
Giải các phương trình, bất phương trình sau:
1) \(\sqrt{3x+7}-5< 0\)
2) \(\sqrt{-2x-1}-3>0\)
3) \(\dfrac{\sqrt{3x-2}}{6}-3=0\)
4) \(-5\sqrt{-x-2}-1< 0\)
5) \(-\dfrac{2}{3}\sqrt{-3-x}-3>0\)
1) \(\sqrt[]{3x+7}-5< 0\)
\(\Leftrightarrow\sqrt[]{3x+7}< 5\)
\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)
\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)
\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)
Gỉai bất phương trình
\(4\sqrt{x^2-4x+1}-\sqrt{x^4+2x^3+x^2-96x}\le x^2-3x-4\)
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) \(\sqrt[3]{x+6}-2\sqrt{x-1}=4-x^2\)
Giải phương trình
\(-3x^2+x+3+\left(\sqrt{3x+2}-4\right)\sqrt{3x-2x^2}+\left(x-1\right)\sqrt{3x+2}=0\)