\(\frac{x^2}{5}\)- \(\frac{2x}{3}\)=\(\frac{x+5}{6}\)
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WR
11 tháng 6 2017
1)
a) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}< =>\frac{2\left(x+5\right)}{2\left(3x-6\right)}-\frac{3x-6}{2\left(3x-6\right)}=\frac{3\left(2x-3\right)}{3\left(2x-4\right)}.\)
(đk:x khác \(\frac{1}{2}\))
\(\frac{2x+10}{6x-12}-\frac{3x-6}{6x-12}=\frac{6x-9}{6x-12}< =>2x+10-3x+6=6x-9< =>x=\frac{25}{7}\)
Vậy x=\(\frac{25}{7}\)
b) /7-2x/=x-3 \(x\ge\frac{7}{2}\)
(đk \(x\ge3,\frac{7}{2}< =>x\ge\frac{7}{2}\))
\(\Rightarrow\orbr{\begin{cases}7-2x=x-3\\7-2x=-\left(x-3\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\left(< \frac{7}{2}\Rightarrow l\right)\\x=4\left(tm\right)\end{cases}}}\)
Vậy x=4
2)
\(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}>\frac{x-4}{5}+\frac{x-5}{6}\)
\(\Leftrightarrow\frac{30\left(x-1\right)}{60}+\frac{20\left(x-2\right)}{60}+\frac{15\left(x-3\right)}{60}-\frac{12\left(x-4\right)}{60}-\frac{10\left(x-5\right)}{60}>0\)
\(\Leftrightarrow30x-30+20x-40+15x-45-12x+48-10x+50>0\Leftrightarrow43x-17>0\Leftrightarrow x>\frac{17}{43}\)
HQ
Hà Quang Minh
Giáo viên
13 tháng 9 2023
\(\frac{x}{3} + \frac{{2x + 1}}{6} = \frac{{4\left( {x - 2} \right)}}{5}\)
\(\frac{{10x}}{{3.10}} + \frac{{\left( {2x + 1} \right).5}}{{6.5}} = \frac{{6.4\left( {x - 2} \right)}}{{5.6}}\)
\(\frac{{10x}}{{30}} + \frac{{10x + 5}}{{30}} = \frac{{24x - 48}}{{30}}\)
\(10x + 10x + 5 = 24x - 48\)
\(10x + 10x - 24x = - 5 - 48\)
\( - 4x = - 53\)
\(x = \left( { - 53} \right):\left( { - 4} \right)\)
\(x = \frac{{53}}{4}\)
Vậy phương trình có nghiệm là \(x = \frac{{53}}{4}\).
20 tháng 7 2019
\(\text{a) }\frac{6}{x-4}-\frac{x}{x+2}=\frac{6}{x-4}.\frac{x}{x+2}\)
\(ĐKXĐ:x\ne-2;x\ne4\)
\(MTC:\left(x-4\right)\left(x+2\right)\)
\(\Leftrightarrow\frac{6\left(x+2\right)}{\left(x-4\right)\left(x+2\right)}-\frac{x\left(x-4\right)}{\left(x-4\right)\left(x+2\right)}=\frac{6x}{\left(x-4\right)\left(x+2\right)}\)
\(\Rightarrow6\left(x+2\right)-x\left(x-4\right)=6x\)
\(\Leftrightarrow6x+12-x^2+4x=6x\)
\(\Leftrightarrow6x+12-x^2+4x-6x=0\)
\(\Leftrightarrow-x^2+4x+12=0\)
\(\Leftrightarrow-\left(x^2-4x-12\right)=0\)
\(\Leftrightarrow x^2-4x-12=0\)
\(\Leftrightarrow x^2+2x-6x-12=0\)
\(\Leftrightarrow x\left(x+2\right)-6\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-6\right)=0\)
\(\Leftrightarrow x=-2\left(\text{loại}\right)\text{ hoặc }x=6\left(\text{nhận}\right)\)
Vậy \(S=\left\{6\right\}\)
\(\text{b) }\frac{2x+3}{2x-1}=\frac{x-3}{x+5}\)
\(ĐKXĐ:x\ne\frac{1}{2};x\ne-5\)
\(\Leftrightarrow\left(2x+3\right)\left(x+5\right)=\left(2x-1\right)\left(x-3\right)\left[\text{Tỉ lệ thức}\right]\)
\(\Leftrightarrow2x^2+10x+3x+15=2x^2-6x-x+3\)
\(\Leftrightarrow2x^2+13x+15=2x^2-7x+3\)
\(\Leftrightarrow2x^2+13x-2x^2+7x=3-15\)
\(\Leftrightarrow20x=-12\)
\(\Leftrightarrow x=\frac{-12}{20}=\frac{-3}{5}\)
Vậy \(S=\left\{\frac{-3}{5}\right\}\)
2 tháng 1 2017
\(\frac{6-2x}{\sqrt{5-x}}+\frac{6+2x}{\sqrt{5+x}}=\frac{8}{3}\)\(\frac{\left(6-2x\right)\left(\sqrt{5+x}\right)}{\left(\sqrt{5+x}\right)\left(\sqrt{5-x}\right)}-\frac{\left(6+2x\right)\left(\sqrt{5-x}\right)}{\left(\sqrt{5+x}\right)\left(\sqrt{5-x}\right)}=\frac{8\left(\sqrt{5+x}\right)\left(\sqrt{5-x}\right)}{3\left(\sqrt{5+x}\right)\left(\sqrt{5-x}\right)}\)
\(3\left(6-2x\right)\left(\sqrt{5+x}\right)-3\left(6+2x\right)\left(\sqrt{5-x}\right)=8\left(\sqrt{5+x}\right)\left(\sqrt{5-x}\right)\)
29 tháng 6 2020
ĐK: \(-5< x< 5\)
Đặt \(a=\sqrt{5+x};b=\sqrt{5-x}\left(a,b>0\right)\)
Khi đó ta có \(6-2x=2b^2-4;6+2x=2a^2-4\)
Khi đó ta có:
\(\frac{2b^2-4}{a}+\frac{2a^2-4}{b}=\frac{8}{3}\Leftrightarrow\left(2b^2-4\right)a+\left(2a^2-4\right)b=\frac{8}{3}ab\)
\(\Leftrightarrow2ab\left(a+b\right)-4\left(a+b\right)=\frac{8}{3}ab\)
Từ đó ta có hệ phương trình
\(\hept{\begin{cases}2ab\left(a+b\right)-4\left(a+b\right)=\frac{8}{3}ab\\a^2+b^2=10\end{cases}\Leftrightarrow\hept{\begin{cases}2ab\left(a+b\right)-4\left(a+b\right)=\frac{8}{3}ab\\\left(a+b\right)^2-2ab=10\end{cases}}}\)
Đặt S=a+b; P=ab (\(S\ge\sqrt{10}\))
Hệ phương trình trở thành
\(\hept{\begin{cases}2SP-4S=\frac{8}{3}P\left(1\right)\\S^2-2P=10\left(2\right)\end{cases}}\)
Từ phương trình (2) ta có \(P=\frac{S^2-10}{2}\)thế lên phương trình trên và rút gọn ta được \(6S^3-8S^2-84S+80=0\Leftrightarrow\left(S-4\right)\left(3S^2+8S-10\right)=0\Leftrightarrow S=4\left(tmđk\right)\)
\(3S^2+8S-10=0\left(VN\right)\)vì \(S>\sqrt{10}\)
S=4 \(\Rightarrow P=3\Leftrightarrow\sqrt{5+x}\sqrt{5-b}=3\Leftrightarrow25-x^2=9\Leftrightarrow x^2=16\Leftrightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}\left(tm\right)}\)
Vậy PT có 2 nghiệm là x=4; x=-4
MSC 90
TA CO ...............................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................................
\(\frac{x^2}{5}-\frac{2x}{3}=\frac{x+5}{6}\)
\(\Leftrightarrow\frac{6x^2}{30}-\frac{20x}{30}=\frac{5x+25}{30}\)
\(\Leftrightarrow6x^2-20x=5x+25\)
\(\Leftrightarrow6x^2-25x-25=0\)
Giải phương trình ra oy tìm x ok