giải phương trình
a)\(\sqrt[3]{x+2}+\sqrt[3]{x-2}=\sqrt[3]{5x}\)
b) \(2\sqrt[3]{\left(x+2\right)^2}-\sqrt[3]{\left(x-2\right)^2}=\sqrt[3]{x^2-4}\)
Giải các phương trình
a) \(2\sqrt{3}x^2+x+1=\sqrt{3}\left(x+1\right)\)
b) \(5x^2-3x+1=2x+31\)
c) \(x^2+2\sqrt{2}x+4=3\left(x+\sqrt{2}\right)\)
a: \(x^2\cdot2\sqrt{3}+x+1=\sqrt{3}\cdot\left(x+1\right)\)
=>\(x^2\cdot2\sqrt{3}+x\left(1-\sqrt{3}\right)+1-\sqrt{3}=0\)
\(\text{Δ}=\left(1-\sqrt{3}\right)^2-4\cdot2\sqrt{3}\left(1-\sqrt{3}\right)\)
\(=4-2\sqrt{3}-8\sqrt{3}\left(1-\sqrt{3}\right)\)
\(=4-2\sqrt{3}-8\sqrt{3}+24=28-10\sqrt{3}=\left(5-\sqrt{3}\right)^2>0\)
Do đó: Phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x_1=\dfrac{-\left(1-\sqrt{3}\right)-\left(5-\sqrt{3}\right)}{2\cdot2\sqrt{3}}=\dfrac{-1+\sqrt{3}-5+\sqrt{3}}{4\sqrt{3}}=\dfrac{1-\sqrt{3}}{2}\\x_2=\dfrac{-\left(1-\sqrt{3}\right)+5-\sqrt{3}}{2\cdot2\sqrt{3}}=\dfrac{4}{4\sqrt{3}}=\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)
b: \(5x^2-3x+1=2x+31\)
=>\(5x^2-3x+1-2x-31=0\)
=>\(5x^2-5x-30=0\)
=>\(x^2-x-6=0\)
=>(x-3)(x+2)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
c: \(x^2+2\sqrt{2}x+4=3\left(x+\sqrt{2}\right)\)
=>\(x^2+2\sqrt{2}x+4-3x-3\sqrt{2}=0\)
=>\(x^2+x\left(2\sqrt{2}-3\right)+4-3\sqrt{2}=0\)
\(\text{Δ}=\left(2\sqrt{2}-3\right)^2-4\left(4-3\sqrt{2}\right)\)
\(=17-12\sqrt{2}-16+12\sqrt{2}=1\)>0
Do đó, phương trình có hai nghiệm phân biệt là:
\(\left[{}\begin{matrix}x_1=\dfrac{-\left(2\sqrt{2}-3\right)-1}{2}=\dfrac{-2\sqrt{2}+3-1}{2}=-\sqrt{2}+1\\x_2=\dfrac{-\left(2\sqrt{2}-3\right)+1}{2}=\dfrac{-2\sqrt{2}+4}{2}=-\sqrt{2}+2\end{matrix}\right.\)
Giải phương trình
a) \(\left(\sqrt{1+x}+\sqrt{1-x}\right)\left(2+2\sqrt{1-x^2}\right)=8\)
b) \(\sqrt{2x+3}+\sqrt{x+1}=3x+2\sqrt{2x^2+5x+3}-16\)
a. ĐKXĐ: \(-1\le x\le1\)
Đặt \(\sqrt{1+x}+\sqrt{1-x}=t>0\)
\(\Rightarrow t^2=2+2\sqrt{1-t^2}\)
Pt trở thành:
\(t.t^2=8\Leftrightarrow t^3=8\Leftrightarrow t=2\)
\(\Rightarrow\sqrt{1+x}+\sqrt{1-x}=2\)
\(\Leftrightarrow2+2\sqrt{1-x^2}=2\)
\(\Leftrightarrow1-x^2=0\Rightarrow x=\pm1\)
b.
ĐKXĐ: \(x\ge-1\)
Đặt \(\sqrt{2x+3}+\sqrt{x+1}=t>0\)
\(\Rightarrow t^2=3x+4+2\sqrt{2x^2+5x+3}\)
Pt trở thành:
\(t=t^2-4-16\Leftrightarrow...\)
giải các phương trình
a \(\sqrt{7+\sqrt{2x}=3+\sqrt{5}}\)
b \(\sqrt{3x^2-4x}=2x-3\)
c\(\dfrac{\left(7-x\right)\sqrt{7-x}+\left(x-5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}=2\)
a) \(\sqrt{7+\sqrt{2x}=3+\sqrt{5}}\) (x≥0) Đặt \(\sqrt{2x}\) = a ( a>0 )
Khi đó pt :
<=> 7+a =3 + \(\sqrt{5}\)
<=> 4+a = \(\sqrt{5}\)
<=> (4+a)\(^2\) = 5
<=> 16 + 8a + a\(^2\) = 5
<=>a\(^2\) + 8a+ 11 = 0
<=> a = -4 + \(\sqrt{5}\) (Loại) và a = -4-\(\sqrt{5}\)(Loại)
Vậy Pt vô nghiệm.
b) \(\sqrt{3x^2-4x}\) = 2x-3
<=> 3x\(^2\)- 4x = 4x\(^2\)-12x + 9
<=> x\(^2\)-8x+9 = 0
<=> x=1 , x=9
Vậy S={1;9}
c\(\dfrac{\left(7-x\right)\sqrt{7-x}+\left(x-5\right)\sqrt{x-5}}{\sqrt{7-x}+\sqrt{x-5}}\) = 2
<=> \(\dfrac{\left(\sqrt{7-x}\right)^3+\left(\sqrt{x-5}\right)^3}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\dfrac{\left(\sqrt{7-x}+\sqrt{x-5}\right)\left(7-x-\sqrt{\left(7-x\right)\left(x-5\right)}+x-5\right)}{\sqrt{7-x}+\sqrt{x-5}}=2\)
<=> \(\sqrt{\left(7-x\right)\left(x-5\right)}=0\)
<=> x=7,x=5
Vậy x=5 hoặc x=7
giải pt:
a, \(\sqrt[3]{64-5x}+\sqrt[3]{18+5x}=4\)
b, \(\sqrt{1+\sqrt{1-x^2}}\left(\sqrt{\left(1+x\right)^3}-\sqrt{\left(1-x\right)^3}\right)=2+\sqrt{1-x^2}\)
c, \(x^2-2x-3=\sqrt{x+3}\)
Giải phương trình:
a) \(5x^2-10x=4\left(x-1\right)\sqrt{x^2-2x+2}\)
b) \(\sqrt{2x^2+22x+29}-x-2=2\sqrt{2x+3}\)
c) \(x^3-7x^2+9x+12=\left(x-3\right)\left(x-2+5\sqrt{x-3}\right)\left(\sqrt{x-3}-1\right)\)
Giải phương trình:
a) \(5x^2-10x=4\left(x-1\right)\sqrt{x^2-2x+2}\)
b) \(\sqrt{2x^2+22x+29}-x-2=2\sqrt{2x+3}\)
c) \(x^3-7x^2+9x+12=\left(x-3\right)\left(x-2+5\sqrt{x-3}\right)\left(\sqrt{x-3}-1\right)\)
Mọi người giúp gấp với ạ.
Giải các bất phương trình, hệ phương trình
a) \(\dfrac{x^2\left(3x-2\right)\left(x^2-1\right)}{\left(-x^2+2x-3\right)\left(2-x\right)^2}\ge0\)
b) \(\dfrac{x-5}{x-1}>2\)
c) \(2x-\sqrt{x^2-5x-14}< 1\)
d) \(x+\sqrt{x^2-4x-5}< 4\)
e) \(\left\{{}\begin{matrix}\left(4-x\right)\left(x^2-2x-3\right)< 0\\x^2\ge\left(x^2-x-3\right)^2\end{matrix}\right.\)
* giải phương trình
a. \(\sqrt{\left(x-3\right)^2}=2\)
b.\(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
a) Ta có: \(\sqrt{\left(x-3\right)^2}=2\)
\(\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
b) ĐKXĐ: \(x\ge-2\)
Ta có: \(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)
\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+\dfrac{4}{5}\cdot5\sqrt{x+2}=6\)
\(\Leftrightarrow2\sqrt{x+2}=6\)
\(\Leftrightarrow x+2=9\)
hay x=7(thỏa ĐK)
a) \(\Leftrightarrow\left|x-3\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=2\\x-3=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\)
Vậy:.....
b) ĐKXĐ: x ≥ -2
\(\Leftrightarrow\sqrt{9}.\sqrt{x+2}-5.\sqrt{x+2}+\dfrac{4}{5}.\sqrt{25}.\sqrt{x+2}=6\)
<=> \(\sqrt{x+2}.\left(3-5+\dfrac{4}{5}.5\right)=6\)
\(\Leftrightarrow2.\sqrt{x+2}=6\)
\(\Leftrightarrow\sqrt{x+2}=3\)
<=> x + 2 = 9
<=> x = 7
Tham khảo ạ:
a) Ta có: √(x−3)2=2(x−3)2=2
⇔|x−3|=2⇔|x−3|=2
⇔[x−3=2x−3=−2⇔[x=5x=1⇔[x−3=2x−3=−2⇔[x=5x=1
b) ĐKXĐ: x≥−2x≥−2
Ta có: ⇔3√x+2−5√x+2+45⋅5√x+2=6⇔3x+2−5x+2+45⋅5x+2=6
⇔2√x+2=6⇔2x+2=6
⇔x+2=9⇔x+2=9
hay x=7(thỏa ĐK)
a) Giải phương trình trên tập số thực:
\(x^3-4x^2-5x+6=\sqrt[3]{7x^2+9x-4}\)
b) Giải hệ phương trình sau:
\(\left\{{}\begin{matrix}x^2+2x\sqrt{xy}=y^2\sqrt{y}\\\left(4x^3+y^3+3x^2\sqrt{x}\right)\left(15\sqrt{x}+y\right)=3\sqrt{x}\left(y\sqrt{y}+x\sqrt{y}+4x\sqrt{x}\right)^2\end{matrix}\right.\) ; với \(x,y\inℝ\)
a) \(x^3-4x^2-5x+6=\sqrt[3]{7x^2+9x-4}\)
\(\Leftrightarrow-7x^2-9x+4+x^3+3x^2+4x+2=\sqrt[3]{7x^2+9x-4}\)
\(\Leftrightarrow-\left(7x^2+9x-4\right)+\left(x+1\right)^3+x+1=\sqrt[3]{7x^2+9x-4}\) (*)
Đặt \(\sqrt[3]{7x^2+9x-4}=a;x+1=b\)
Khi đó (*) \(\Leftrightarrow-a^3+b^3+b=a\)
\(\Leftrightarrow\left(b-a\right).\left(b^2+ab+a^2+1\right)=0\)
\(\Leftrightarrow b=a\)
Hay \(x+1=\sqrt[3]{7x^2+9x-4}\)
\(\Leftrightarrow\left(x+1\right)^3=7x^2+9x-4\)
\(\Leftrightarrow x^3-4x^2-6x+5=0\)
\(\Leftrightarrow x^3-4x^2-5x-x+5=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2+x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-1\pm\sqrt{5}}{2}\end{matrix}\right.\)
Giải Phương Trình
\(\sqrt{\left(2x+3\right)^2}=5\)
\(\sqrt{9\left(x-2\right)^2}=18\)
\(\sqrt{9x-18}-\sqrt{4x-8}+3\sqrt{x-2}=40\)
\(\sqrt{4.\left(x-3\right)^2}=8\)
\(\sqrt{5x-6}-3=0\)