giai phuong trinh
\(18x^2-2x-\frac{17}{3}+9\sqrt{x-\frac{1}{3}}=0\)
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P
1
2 tháng 10 2020
ĐK: \(x\ge\frac{1}{3}\)
Pt đã cho tương đương với \(\left(18x^2-2x-\frac{8}{3}\right)+9\left(\sqrt{x-\frac{1}{3}}-\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left(18x-8\right)\left(x+\frac{1}{3}\right)+9\frac{x-\frac{1}{3}-\frac{1}{9}}{\sqrt{x-\frac{1}{3}}+\frac{1}{3}}=0\)
\(\Leftrightarrow\left(x-\frac{4}{9}\right)\text{[}18\left(x+\frac{1}{3}\right)+9\frac{1}{\sqrt{x-\frac{1}{3}}+\frac{1}{2}}\text{]}=0\Rightarrow x=\frac{4}{9}\)
CM: Với \(x\ge\frac{1}{3}\Rightarrow18\left(x+\frac{1}{3}\right)+9\frac{1}{\sqrt{x-\frac{1}{3}}+\frac{1}{3}}>0\)
Pt đã cho có nghiệm \(x=\frac{4}{9}\)
25 tháng 6 2018
\(\Rightarrow\frac{2}{x^2+x+3x+3}+\frac{5}{x^2+3x+8x+24}+\frac{2}{x^2+10x+8x+80}=\frac{9}{52}\)
\(\Rightarrow\frac{2}{x\left(x+1\right)+3\left(x+1\right)}+\frac{5}{x\left(x+3\right)+8\left(x+3\right)}+\frac{2}{x\left(x+10\right)+8\left(x+10\right)}=\frac{9}{52}\)
\(\Rightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)
\(\Rightarrow\frac{1}{x+1}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+10}=\frac{9}{52}\)
\(\Rightarrow\frac{1}{x+1}-\frac{1}{x+10}=\frac{9}{52}\Rightarrow\frac{x+10-x-1}{\left(x+1\right)\left(x+10\right)}=\frac{9}{52}\Rightarrow\frac{9}{x^2+11x+10}=\frac{9}{52}\)
\(\Rightarrow x^2+11x+10=52\Rightarrow x^2+2\cdot\frac{11}{2}x+\frac{121}{4}-\frac{81}{4}=52\)
\(\Rightarrow\left(x+\frac{11}{2}\right)^2=\frac{289}{4}\Rightarrow x+\frac{11}{2}=\frac{17}{2}\Rightarrow x=\frac{17}{2}-\frac{11}{2}=\frac{6}{2}=3\Rightarrow x=3\)
25 tháng 6 2018
\(\frac{2}{x^2+4x+3}+\frac{5}{x^2+11x+24}+\frac{2}{x^2+18x+80}=\frac{9}{52}\)(ĐKXĐ: x khác -1;-3;-8;-10)
\(\Leftrightarrow\frac{2}{x^2+x+3x+3}+\frac{5}{x^2+3x+8x+24}+\frac{2}{x^2+8x+10x+80}=\frac{9}{52}\)
\(\Leftrightarrow\frac{2}{\left(x+1\right)\left(x+3\right)}+\frac{5}{\left(x+3\right)\left(x+8\right)}+\frac{2}{\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)
\(\Leftrightarrow\frac{2\left(x+8\right)\left(x+10\right)+5\left(x+1\right)\left(x+10\right)+2\left(x+1\right)\left(x+3\right)}{\left(x+1\right)\left(x+3\right)\left(x+8\right)\left(x+10\right)}=\frac{9}{52}\)
\(\Leftrightarrow\frac{9x^2+99x+216}{x^4+22x^3+155x^2+374x+240}=\frac{9}{52}\)
\(\Rightarrow468x^2+5148x+11232=9x^4+198x^3+1395x^2+3366x+2160\)
\(\Leftrightarrow9x^4+198x^3+927x^2-1782x-9072=0\)
\(\Leftrightarrow x^4+22x^3+103x^2-198x-1008=0\)
\(\Leftrightarrow x^4-3x^3+25x^3-75x^2+178x^2-534x+336x-1008=0\)
\(\Leftrightarrow x^3\left(x-3\right)+25x^2\left(x-3\right)+178x\left(x-3\right)+336\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^3+25x^2+178x+336\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^3+3x^2+22x^2+66x+112x+336\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left[x^2\left(x+3\right)+22x\left(x+3\right)+112\left(x+3\right)\right]=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+22x+112\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x^2+8x+14x+112\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)\left[x\left(x+8\right)+14\left(x+8\right)\right]=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3\right)\left(x+8\right)\left(x+14\right)=0\)
\(\Leftrightarrow\frac{\orbr{\begin{cases}x+3=0\\x-3=0\end{cases}}}{\orbr{\begin{cases}x+8=0\\x+14=0\end{cases}}}\Leftrightarrow\frac{\orbr{\begin{cases}x=-3\left(\times\right)\\x=3\end{cases}}}{\orbr{\begin{cases}x=-8\left(\times\right)\\x=-14\end{cases}}}\)(Vì x=-3 và x=-8 không t/m ĐKXĐ)
Vậy tập nghiệm của pt là \(S=\left\{3;-14\right\}.\)
XO
20 tháng 2 2020
b) \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)
=> \(\left(\frac{x-90}{10}-1\right)+\left(\frac{x-76}{12}-2\right)+\left(\frac{x-58}{14}-3\right)+\left(\frac{x-36}{16}-4\right)+\left(\frac{x-15}{17}-5\right)=0\)
=> \(\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)
=> \(\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)
Vì \(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\ne0\)
=> x - 100 = 0
=> x = 100
PH
0
TT
30 tháng 7 2015
Đkiện: x <1 hoặc x \(\ge\frac{3}{2}\)
\(\sqrt{\frac{2x-3}{x-1}}=2\) (1)
(1) => \(\frac{2x-3}{x-1}=4\)
=> 2x - 3 = 4x - 4
<=> 2x - 4x = -4 + 3
<=> -2x = -1
<=> x = \(\frac{1}{2}\)( TMĐK)
Vậy x = \(\frac{1}{2}\)
b, Đkiện: x \(\ge\frac{3}{2}\)
(1) => \(\sqrt{2x-3}=2\sqrt{x-1}\)
=>2x - 3 = 4(x - 1)
<=> 2x -3 = 4x -4
<=> -2x = -1
<=> x = \(\frac{1}{2}\)(ko TMĐK)
Vậy pt vô nghiệm
T
4
10 tháng 8 2019
Pt tương đương:
\(\sqrt[3]{4x-3}\)-\(\sqrt[3]{3x+1}\)=\(\sqrt[3]{5-x}\)+\(\sqrt[3]{2x-9}\)
\(\Leftrightarrow\)-3\(\sqrt[3]{\text{(4x-3)(3x+1)}}\)(\(\sqrt[3]{4x-3}\)-\(\sqrt[3]{3x+1}\))=3\(\sqrt[3]{\left(5-x\right)\left(2x-9\right)}\)(\(\sqrt[3]{5-x}\)+\(\sqrt[3]{2x-9}\))
\(\Leftrightarrow\)\(\orbr{\begin{cases}\sqrt[3]{4x-3}-\sqrt[3]{3x+1}=\sqrt[3]{5-x}+\sqrt[3]{2x-9}=0\left(1\right)\\3\sqrt[3]{-12x^2+5x+3}=3\sqrt[3]{-2x^2+19x-45}\left(2\right)\end{cases}}\)
(1)<=>4x-3=3x+1 và x-5=2x-9<=>x=4
(2)<=>-12x2+5x+3=-2x2+19x-45<=>-5x2-7x+24=0<=>x=8/5 và x=-3
bạn thử các giá trị x=4,x=8/5 và x=-3 vào pt và kết luận
UI
10 tháng 8 2019
mik ko hieu vi sao ban suy ra duoc (1) va (2)
bn co the viet ro ra duoc ko ?
theo mik thay thi 2 pt do dau co tuong duong
DT
3
NV
20 tháng 5 2017
ĐKXĐ: \(x\ge\frac{1}{2}\)
Đề \(\Rightarrow\sqrt{\frac{x+7}{x+1}}-\sqrt{3}+8-2x^2-\left(\sqrt{2x-1}-\sqrt{3}\right)=0\)
Nhân liên hợp ta được:
\(\frac{\left(\sqrt{\frac{x+7}{x+1}}-\sqrt{3}\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{\left(\sqrt{2x-1}-\sqrt{3}\right)\left(\sqrt{2x+1}+\sqrt{3}\right)}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\frac{\frac{x+7}{x+1}-3}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(4-x^2\right)-\frac{2x-1-3}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\frac{\frac{-2x+4}{x+1}}{\sqrt{\frac{x+7}{x+1}}+\sqrt{3}}+2\left(2-x\right)\left(2+x\right)-\frac{2x-4}{\sqrt{2x+1}+\sqrt{3}}=0\)
\(\Rightarrow\left(x-2\right)\left[\frac{-2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}\right]=0\)
mà \(-\frac{2}{\left(x+1\right)\left(\sqrt{\frac{x+7}{x+1}}+\sqrt{3}\right)}-2\left(2+x\right)-\frac{2}{\sqrt{2x+1}+\sqrt{3}}< 0\)
=> x - 2 = 0 => x = 2
Vậy x = 2
HL
0
\(18x^2-2x-\frac{17}{3}+9\sqrt{x-\frac{1}{3}}=0\)
Điều kiện: \(x\ge\frac{1}{3}\)
Đặt \(\sqrt{x-\frac{1}{3}}=a\left(a\ge0\right)\)
\(\Rightarrow x=a^2+\frac{1}{3}\)
Ta suy ra phương trình tương đương với
\(18\left(a^2+\frac{1}{3}\right)^2-2\left(a^2+\frac{1}{3}\right)-\frac{17}{3}+9a=0\)
\(\Leftrightarrow54a^4+30a^2+27a-13=0\)
\(\Leftrightarrow\left(3a-1\right)\left(18a^3+6a^2+12a+13\right)=0\)
Dễ thấy \(18a^3+6a^2+12a+13>0\) vì \(a\ge0\)
\(\Rightarrow3a-1=0\)
\(\Leftrightarrow a=\frac{1}{3}\)
\(\Leftrightarrow\sqrt{x-\frac{1}{3}}=\frac{1}{3}\)
\(\Leftrightarrow x-\frac{1}{3}=\frac{1}{9}\)
\(\Leftrightarrow x=\frac{4}{9}\)