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Em sửa tí:
"Tức là \(-\left(8x^2+20x+21-\frac{x+6}{2}\right)< 0\) (2)" Đánh số (2) ở chỗ này mới đúng nha!
ĐK \(x\ge-3\)
Pt
<=> \(\left(x+3\right)\sqrt{x+3}+3\left(x+3\right)+3\sqrt{x+3}+1=8x^3+12x^2+6x+1\)
<=> \(\left(\sqrt{x+3}+1\right)^3=\left(2x+1\right)^3\)
=> \(\sqrt{x+3}=2x\)
=> \(\left\{{}\begin{matrix}x\ge0\\4x^2-x-3=0\end{matrix}\right.\)
=> \(x=1\)(tmĐK)
Vậy x=1
Bài 6:
ĐK: $x\geq \frac{2}{3}$
Đặt $\sqrt{4x+1}=a; \sqrt{3x-2}=b(a,b\geq 0)$
PT trở thành:
$a-b=a^2-b^2$
$\Leftrightarrow (a-b)(a+b)-(a-b)=0$
$\Leftrightarrow (a-b)(a+b-1)=0$
Nếu $a-b=0\Leftrightarrow 4x+1=3x-2\Leftrightarrow x=-3$ (loại vì không thỏa ĐKXĐ)
Nếu $a+b-1=0$
$\Leftrightarrow b=1-a$
$\Leftrightarrow \sqrt{3x-2}=1-\sqrt{4x+1}$
$\Rightarrow 3x-2=4x+2-2\sqrt{4x+1}$
$\Leftrightarrow x+4=2\sqrt{4x+1}$
$\Rightarrow (x+4)^2=4(4x+1)$
$\Leftrightarrow x^2-8x+12=0\Leftrightarrow x=6$ hoặc $x=2$
Vậy.......
Bài 5:
ĐK: $x\geq -2$
PT $\Leftrightarrow 3\sqrt{(x+2)(x^2-2x+4)}=2x^2-3x+10$
Đặt $\sqrt{x+2}=a; \sqrt{x^2-2x+4}=b(a,b\geq 0)$
Khi đó PT trở thành:
$3ab=2b^2+a^2$
$\Leftrightarrow a^2-3ab+2b^2=0$
$\Leftrightarrow a(a-b)-2b(a-b)=0$
$\Leftrightarrow (a-b)(a-2b)=0$
Nếu $a-b=0\Rightarrow a^2-b^2=0$
$\Leftrightarrow x+2-(x^2-2x+4)=0$
$\Leftrightarrow x^2-3x+2=0\Rightarrow x=1$ hoặc $x=2$ (thỏa mãn)
Nếu $a-2b=0\Rightarrow 4b^2-a^2=0$
$\Leftrightarrow 4(x^2-2x+4)-(x+2)=0$
$\Leftrightarrow 4x^2-9x+14=0$ (pt vô nghiệm)
Vậy.........
5.
ĐKXĐ: \(-\frac{1}{2}\le x\le\frac{1}{2}\)
\(\Leftrightarrow\frac{1}{2}-x+\frac{1}{2}+x+2\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=1\)
\(\Leftrightarrow\sqrt{\left(\frac{1}{2}-x\right)\left(\frac{1}{2}+x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
6.
ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x^2-1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)
\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}-\sqrt{x-1}-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\left(vn\right)\end{matrix}\right.\)
2.
ĐKXĐ: \(x\ge-1\)
\(\Leftrightarrow2\left(x^2+2\right)=5\sqrt{\left(x+1\right)\left(x^2-x+1\right)}\)
Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{x^2-x+1}=b>0\end{matrix}\right.\)
\(\Leftrightarrow2\left(a^2+b^2\right)=5ab\)
\(\Leftrightarrow2a^2-5ab+2b^2=0\)
\(\Leftrightarrow\left(a-2b\right)\left(2a-b\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2a=b\\a=2b\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x+1}=\sqrt{x^2-x+1}\\\sqrt{x+1}=2\sqrt{x^2-x+1}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+4=x^2-x+1\\x+1=4x^2-4x+4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-3=0\\4x^2-5x+3=0\end{matrix}\right.\) \(\Leftrightarrow...\)
mọi người giúp mình với ạ,mai mình phải nộp rồi nhưng kô biết làm .Mong mn giúp đỡ!!!
7/
ĐKXĐ: \(-3\le x\le\frac{2}{3}\)
\(\Leftrightarrow2x+8\sqrt{x+3}+4\sqrt{3-2x}=2\)
\(\Leftrightarrow8\sqrt{x+3}+4\sqrt{3-2x}-\left(3-2x\right)+1=0\)
\(\Leftrightarrow8\sqrt{x+3}+\sqrt{3-2x}\left(4-\sqrt{3-2x}\right)+1=0\)
Do \(x\ge-3\Rightarrow3-2x\le9\Rightarrow\sqrt{3-2x}\le3\)
\(\Rightarrow4-\sqrt{3-2x}>0\)
\(\Rightarrow VT>0\)
Phương trình vô nghiệm (bạn coi lại đề)
5/
\(\Leftrightarrow8x^2-3x+6-4x\sqrt{3x^2+x+2}=0\)
\(\Leftrightarrow\left(4x^2-4x\sqrt{3x^2+x+2}+3x^2+x+2\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(2x-\sqrt{3x^2+x+2}\right)^2+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-\sqrt{3x^2+x+2}=0\\x-2=0\end{matrix}\right.\) \(\Rightarrow x=2\)
6/
ĐKXĐ: ....
\(\Leftrightarrow\left(x-2000-2\sqrt{x-2000}+1\right)+\left(y-2001-2\sqrt{y-2001}+1\right)+\left(z-2002-2\sqrt{z-2002}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2000}-1\right)^2+\left(\sqrt{y-2001}-1\right)^2+\left(\sqrt{z-2002}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-2000}-1=0\\\sqrt{y-2001}-1=0\\\sqrt{z-2002}-1=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=2001\\y=2002\\z=2003\end{matrix}\right.\)
c, ĐKXĐ: \(x\ge\frac{1}{2}\)
\(\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)
\(\Leftrightarrow\sqrt{2x-2\sqrt{2x-1}}=2\)
\(\Leftrightarrow\sqrt{2x-1-2\sqrt{2x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}-1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|=2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-1}-1=2\\\sqrt{2x-1}-1=-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{2x-1}=3\\\sqrt{2x-1}=-1\left(vn\right)\end{matrix}\right.\)
\(\sqrt{2x-1}=3\Leftrightarrow2x-1=9\Leftrightarrow x=5\left(tm\right)\)
a, ĐKXĐ: \(x\in R\)
\(\sqrt{3x^2}=x+2\)
\(\Leftrightarrow\sqrt{3}\left|x\right|=x+2\)
TH1: \(\sqrt{3}x=x+2\)
\(\Leftrightarrow\left(\sqrt{3}-1\right)x=2\)
\(\Leftrightarrow x=\sqrt{3}+1\)
TH2: \(\sqrt{3}x=-x-2\)
\(\Leftrightarrow\left(\sqrt{3}+1\right)x=-2\)
\(\Leftrightarrow x=1-\sqrt{3}\)
\(x=1+\sqrt[3]{5}+\sqrt[3]{25}\Rightarrow x-1=\sqrt[3]{5}+\sqrt[3]{25}\)
\(\Rightarrow\left(x-1\right)^3=5+25+3.\sqrt[3]{5.25}\left(\sqrt[3]{5}+\sqrt[3]{25}\right)=30+15\left(\sqrt[3]{5}+\sqrt[3]{25}\right)\)
\(\Rightarrow\left(x-1\right)^3=30+15\left(x-1\right)=15+15x\)
Ta có:
\(P=\left(x^3-3x^2+3x-1-15x-14\right)^{10}+2018\)
\(P=\left(\left(x-1\right)^3-15x-14\right)^{10}+2018=\left(15+15x-15x-14\right)^{10}+2018\)
\(\Rightarrow P=1^{10}+2018=1+2018=2019\)