Học tại trường Chưa có thông tin
Đến từ Ninh Bình , Chưa có thông tin
Số lượng câu hỏi 14
Số lượng câu trả lời 748
Điểm GP 136
Điểm SP 566

Người theo dõi (87)

Đang theo dõi (21)

thieu vu
văn tài
Ái Nữ
Trần Thiên Kim
Học 24h

Câu trả lời:

Cách khác:

\(\sqrt[3]{81x-8}=x^3-2x^2+\frac{4}{3}x-2\)

<=> \(3\sqrt[3]{81x-8}=3x^3-6x^2+4x-6\)

<=>\(3\left[\sqrt[3]{81x-8}-\left(3x-2\right)\right]=3x^3-6x^2-5x\)

Xét \(\sqrt[3]{\left(81x-8\right)^2}+\left(3x-2\right)\sqrt[3]{81x-8}+\left(3x-2\right)^2=0\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{81x-8}=a\\3x-2=b\end{matrix}\right.\)

\(a^2+ab+b^2=0\) <=> \(\left(a+\frac{b}{2}\right)^2+\frac{3b^2}{4}=0\)

\(VT\ge0\).Dấu "=" xảy ra <=>\(\left\{{}\begin{matrix}b=0\\a+\frac{b}{2}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}b=0\\a=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}3x-2=0\\\sqrt[3]{81x-8}=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\frac{2}{3}\\x=\frac{8}{81}\end{matrix}\right.\)(L)

=> \(\sqrt[3]{\left(81x-8\right)^2}+\left(3x-2\right)\sqrt[3]{81x-8}+\left(3x-2\right)^2\ne0\)

pt <=>\(\frac{81x-8-\left(3x-2\right)^3}{\sqrt[3]{\left(81x-8\right)^2}+\left(3x-2\right)\sqrt[3]{81x-8}+\left(3x-2\right)^2}=3x^3-6x^2-5x\)

<=>\(\frac{-27x^3+54x^2+45x}{\sqrt[3]{\left(81x-8\right)^2}+\left(3x-2\right)\sqrt[3]{81x-8}+\left(3x-2\right)^2}-\left(3x^3-6x^2-5x\right)=0\)

<=>\(\left(3x^3-6x^2-5x\right)\left(\frac{-9}{\sqrt[3]{\left(81x-8\right)^2}+\left(3x-2\right)\sqrt[3]{81x-8}+\left(3x-2\right)^2}-1\right)=0\)

<=>\(\left[{}\begin{matrix}3x^3-6x^2-5x=0\\-\frac{9}{\sqrt[3]{\left(81x-8\right)^2}+\left(3x-2\right)\sqrt[3]{81x-8}+\left(3x-2\right)^2}=1\end{matrix}\right.\)

<=>\(\left[{}\begin{matrix}x\left(x-\frac{3+2\sqrt{6}}{3}\right)\left(x-\frac{3-2\sqrt{6}}{3}\right)=0\left(1\right)\\-9=\sqrt[3]{\left(81-8\right)^2}+\left(3x-2\right)\sqrt[3]{81x-8}+\left(3x-2\right)^2\left(2\right)\end{matrix}\right.\)

Pt (1) <=> \(\left[{}\begin{matrix}x=0\left(tm\right)\\x=\frac{3+2\sqrt{6}}{3}\left(tm\right)\\x=\frac{3-2\sqrt{6}}{3}\left(tm\right)\end{matrix}\right.\)

Pt (2) vô nghiệm do VP >0

Vậy pt có tập nghiệm \(S=\left\{0,\frac{3+2\sqrt{6}}{3},\frac{3-2\sqrt{6}}{3}\right\}\)

Câu trả lời:

Bấm máy may mắn ra nghiệm đẹp

Đk: \(-1\le x\le\frac{5}{2}\)

PT <=> \(6x^2+20x+2\sqrt{3x+3}=2x^3+52+2\sqrt{5-2x}\)

<=> \(\left[2\sqrt{3x+3}-\left(4+x\right)\right]+6x^2+23x=2x^3+2\left[\sqrt{5-2x}-\left(3-x\right)\right]+54\)

Xét \(-1\le x\) => \(2\sqrt{3x+3}+4+x\ge0+4-1=3>0\)

Xét \(-1\le x\le\frac{5}{2}\) => \(\frac{1}{2}\le\sqrt{5-2x}+3-x\le\sqrt{7}+4\) => \(\sqrt{5-2x}+3-x\ne0\)

Pt <=> \(\frac{4\left(3x+3\right)-\left(4+x\right)^2}{2\sqrt{3x+3}+4+x}+6x^2+23x=2x^3+2.\frac{5-2x-\left(3-x\right)^2}{\sqrt{5-2x}+3-x}+54\)

<=>\(\frac{-x^2+4x-4}{2\sqrt{3x+3}+4x+}-2.\frac{-x^2+4x-4}{\sqrt{5-2x}+3-x}-\left(2x^3-6x^2-23x+54\right)=0\)

<=> \(\frac{-\left(x-2\right)^2}{2\sqrt{3x+3}+4+x}+\frac{2\left(x-2\right)^2}{\sqrt{5-2x}+3-x}-\left(x-2\right)\left(2x^2-2x-27\right)=0\)

<=>\(\left(x-2\right)\left[\frac{-\left(x-2\right)}{2\sqrt{3x+3}+4+x}+\frac{2\left(x-2\right)}{\sqrt{5-2x}+3-x}-2x^2+2x+27\right]=0\)

<=>\(\left[{}\begin{matrix}x-2=0\left(1\right)\\-\frac{\left(x-2\right)}{2\sqrt{3x+3}+4+x}+\frac{2\left(x-2\right)}{\sqrt{5-2x}+3-x}-2x^2+2x+27=0\left(2\right)\end{matrix}\right.\)

Từ (1)=> x=2(t/m pt)

Chắc chắn (2) vô nghiệm nhưng chưa biết CM

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Đau mắt quá thì chuyển qua liên hợp kiểu này đi(dễ hơn)

pt <=> \(\left(\sqrt{3x+3}-3\right)-\left(\sqrt{5-2x}-1\right)+3x^2+10x-x^3-24=0\)

Luôn có \(\left\{{}\begin{matrix}\sqrt{3x+3}+3>0\\\sqrt{5-2x}+1>0\end{matrix}\right.\) với mọi x

pt <=> \(\frac{3x+3-9}{\sqrt{3x+3}+3}-\frac{5-2x-1}{\sqrt{5-2x}+1}-\left(x-2\right)\left(x+3\right)\left(x-4\right)=0\)

<=>\(\frac{3\left(x-2\right)}{\sqrt{3x+3}+3}+\frac{2\left(x-2\right)}{\sqrt{5-2x}+1}-\left(x-2\right)\left(x+3\right)\left(x-4\right)=0\)

<=>\(\left(x-2\right)\left[\frac{3}{\sqrt{3x+3}+3}+\frac{2}{\sqrt{5-2x}+1}-\left(x+3\right)\left(x-4\right)\right]=0\)

<=>\(\left[{}\begin{matrix}x=2\left(tm\right)\\\frac{3}{\sqrt{3x+3}+3}+\frac{2}{\sqrt{5-2x}+1}-\left(x+3\right)\left(x-4\right)=0\left(1\right)\end{matrix}\right.\)

(1) <=>\(\frac{3}{\sqrt{3x+3}+3}+\frac{2}{\sqrt{5-2x}+1}=\left(x+3\right)\left(x-4\right)\)

Tại \(-1\le x\le\frac{5}{2}\)=> \(-10\le\left(x+3\right)\left(x-4\right)\le-\frac{33}{4}< 0\)

=> Vế phải của (1) luôn âm

Xét vế trái của (1) có: \(\left\{{}\begin{matrix}\sqrt{3x+3}+3>0\\\sqrt{5-2x}+1>0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}\frac{3}{\sqrt{3x+3}+3}>0\\\frac{2}{\sqrt{5-2x}+1}>0\end{matrix}\right.\)=> \(\frac{3}{\sqrt{3x+3}+3}+\frac{2}{\sqrt{5-2x}+1}>0\)

=> Vế trái của (1) luôn dương hay (1) vô nghiệm

Vậy pt có 1 nghiệm duy nhất x=2