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Hung nguyen, Trần Thanh Phương, Sky SơnTùng, @tth_new, @Nguyễn Việt Lâm, @Akai Haruma, @No choice teen
help me, pleaseee
Cần gấp lắm ạ!
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1:
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-2\right)=0\)
=>x-3=0 hoặc \(\sqrt{x+3}=2\)
=>x=3 hoặc x+3=4
=>x=1(loại) hoặc x=3(nhận)
2:
\(\Leftrightarrow\left(\sqrt{4x+1}-\sqrt{3x-4}\right)^2=1\)
=>\(4x-1+3x-4-2\sqrt{\left(4x+1\right)\left(3x-4\right)}=1\)
=>\(\sqrt{4\left(4x+1\right)\left(3x-4\right)}=7x-6\)
=>4(12x^2-16x+3x-4)=(7x-6)^2
=>49x^2-84x+36=48x^2-52x-16
=>-84x+36=-52x-16
=>-32x=-52
=>x=13/8
3: =>\(\sqrt{\left(x-5\right)^2}=5-x\)
=>|x-5|=5-x
=>x-5<=0
=>x<=5
4: \(\Leftrightarrow\left|x-4\right|=x+2\)
=>\(\left\{{}\begin{matrix}x>=-2\\\left(x-4\right)^2=\left(x+2\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=-2\\x^2-8x+16=x^2+4x+4\end{matrix}\right.\)
=>x>=-2 và -8x+16=4x+4
=>x=1
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Lời giải:
Trước tiên ta thấy:
\(3(x-1)\sqrt{x^2+x+3}=3x^2+4x+7=2x^2+(x+2)^2+3>0\)
Mà \(3\sqrt{x^2+x+3}\geq 0\)
\(\Rightarrow x-1>0\Rightarrow x>1\)
Phương trình đã cho tương đương với:
\(\Leftrightarrow (x^2+x+3)+(x^2-2x+1)-2(x-1)\sqrt{x^2+x+3}+(x^2+5x+3)-(x-1)\sqrt{x^2+x+3}=0\)
\(\Leftrightarrow [\sqrt{x^2+x+3}-(x-1)]^2+\sqrt{x^2+x+3}[\sqrt{x^2+x+3}-(x-1)]+4x=0(*)\)
Có:
\([\sqrt{x^2+x+3}-(x-1)]^2\geq 0\)
\(\sqrt{x^2+x+3}-(x-1)=\frac{x^2+x+3-(x-1)^2}{\sqrt{x^2+x+3}+x-1}=\frac{3x+2}{\sqrt{x^2+x+3}+x-1}>0, \forall x>1\)
\(4x>0, \forall x>1\)
Do đó: \([\sqrt{x^2+x+3}-(x-1)]^2+\sqrt{x^2+x+3}[\sqrt{x^2+x+3}-(x-1)]+4x>0\) (mâu thuẫn với (*))
Vậy pt vô nghiệm.
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(x\ge\sqrt[3]{7}\)
\(4x^3-x^2+2x-32+\left(x^3-4\right)\left(\sqrt{x^3-7}-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x^2+7x+16\right)+\dfrac{\left(x^3-4\right)\left(x-2\right)\left(x^2+2x+4\right)}{\sqrt{x^3-7}+1}=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x^2+7x+16+\dfrac{\left(x^3-4\right)\left(x^2+2x+4\right)}{\sqrt{x^3-7}+1}\right)=0\)
\(\Leftrightarrow x=2\) (ngoặc đằng sau luôn dương do \(x^3-4=x^3-7+3>0\))
2.
\(\Leftrightarrow\left(2x^3\right)^3+2x^3=x^3+3x^2+3x+1+x+1\)
\(\Leftrightarrow\left(2x^3\right)^3+2x^3=\left(x+1\right)^3+x+1\)
Đặt \(\left\{{}\begin{matrix}2x^3=a\\x+1=b\end{matrix}\right.\)
\(\Rightarrow a^3-b^3+a-b=0\Leftrightarrow\left(a-b\right)\left(a^2+ab+b^2+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Rightarrow2x^3=x+1\Leftrightarrow\left(x-1\right)\left(2x^2+2x+1\right)=0\)
Đặt \(x+1=a,\sqrt{x^2+x+3}=b\left(b>0\right)\)
=> \(a^2+2b^2=x^2+2x+1+2\left(x^2+x+3\right)=3x^2+4x+7\)
Khi đó PT
<=> \(a^2+2b^2-3ab=0\)
<=> \(\orbr{\begin{cases}a=b\\a=2b\end{cases}}\)
+ a=b
=> \(x+1=\sqrt{x^2+x+3}\)
<=>\(\hept{\begin{cases}x\ge-1\\x^2+2x+1=x^2+x+3\end{cases}}\)
=> x=2
+ a=2b
=> \(x+1=2\sqrt{x^2+x+3}\)
<=> \(\hept{\begin{cases}x\ge-1\\x^2+2x+1=4\left(x^2+x+3\right)\end{cases}}\)
=> \(\hept{\begin{cases}x\ge-1\\3x^2+2x+11=0\end{cases}}\)(vô nghiệm )
Vậy x=2
cảm ơn bạn nhiều