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\(\frac{x^2-36}{2x+10}\cdot\frac{3}{6-x}=\frac{\left(x-6\right)\left(x+6\right)}{2x+10}\cdot\frac{3}{6-x}=-\frac{3\left(x+6\right)}{2x+10}=-\frac{3x+18}{2x+10}\)
\(\frac{x^2-4}{x^2-9}\cdot\frac{3x+9}{x+2}=\frac{\left(x-2\right)\left(x+2\right)}{\left(x+3\right)\left(x-3\right)}\cdot\frac{3\left(x+3\right)}{x+2}=\frac{3\left(x-2\right)}{x-3}\)
\(\frac{x^3-8}{5x+20}\cdot\frac{x^2+4x}{x^2+2x+4}=\frac{\left(x-2\right)\left(x^2+2x+4\right)}{5\left(x+4\right)}\cdot\frac{x\left(x+4\right)}{x^2+2x+4}=\frac{x\left(x-2\right)}{5}\)
\(\frac{4x+12}{\left(x+4\right)^2}:\frac{3x+9}{x+4}=\frac{4\left(x+3\right)}{\left(x+4\right)^2}\cdot\frac{x+4}{3\left(x+3\right)}=\frac{4}{3\left(x+4\right)}\)
Ta có : \(\frac{x}{3}=\frac{y}{4}\Rightarrow\frac{x}{9}=\frac{y}{12}\left(1\right)\)
\(\frac{y}{6}=\frac{z}{5}\Rightarrow\frac{y}{12}=\frac{z}{10}\left(2\right)\)
Từ (1) và (2) => \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}\)
Ta có : \(\frac{x}{9}=\frac{y}{12}=\frac{z}{10}=\frac{3x}{27}=\frac{2y}{24}=\frac{5z}{50}=\frac{3x-2y+5z}{27-24+50}=\frac{86}{53}\) (đề sai)
b) Đặt : k = \(\frac{x}{5}=\frac{y}{7}\)
=> k2 \(=\frac{x}{5}.\frac{y}{7}=\frac{xy}{35}=\frac{140}{35}=4\)
=> k = -2;2
+ k = 2 thì \(\frac{x}{5}=2\Rightarrow x=10\)
\(\frac{z}{7}=2\Rightarrow z=14\)
+ k = -2 thì \(\frac{x}{5}=2\Rightarrow x=-10\)
\(\frac{z}{7}=2\Rightarrow z=-14\)
Vậy................................
g) \(\frac{x+2}{98}+\frac{x+4}{96}=\frac{x+6}{94}+\frac{x+8}{92}\)
\(\Leftrightarrow\left(\frac{x+2}{98}+1\right)+\left(\frac{x+4}{96}+1\right)=\left(\frac{x+6}{94}+1\right)+\left(\frac{x+8}{92}+1\right)\)
\(\Leftrightarrow\left(\frac{x+2+98}{98}\right)+\left(\frac{x+4+96}{96}\right)=\left(\frac{x+6+94}{94}\right)+\left(\frac{x+8+92}{92}\right)\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}=\frac{x+100}{94}+\frac{x+100}{92}\)
\(\Leftrightarrow\frac{x+100}{98}+\frac{x+100}{96}-\frac{x+100}{94}-\frac{x+100}{92}=0\)
\(\Leftrightarrow\left(x+100\right).\left(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\right)=0\)
Vì \(\frac{1}{98}+\frac{1}{96}-\frac{1}{94}-\frac{1}{92}\ne0.\)
\(\Leftrightarrow x+100=0\)
\(\Leftrightarrow x=0-100\)
\(\Leftrightarrow x=-100.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{-100\right\}.\)
h) \(\frac{x-12}{77}+\frac{x-11}{78}=\frac{x-74}{15}+\frac{x-73}{16}\)
\(\Leftrightarrow\left(\frac{x-12}{77}-1\right)+\left(\frac{x-11}{78}-1\right)=\left(\frac{x-74}{15}-1\right)+\left(\frac{x-73}{16}-1\right)\)
\(\Leftrightarrow\left(\frac{x-12-77}{77}\right)+\left(\frac{x-11-78}{78}\right)=\left(\frac{x-74-15}{15}\right)+\left(\frac{x-73-16}{16}\right)\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}=\frac{x-89}{15}+\frac{x-89}{16}\)
\(\Leftrightarrow\frac{x-89}{77}+\frac{x-89}{78}-\frac{x-89}{15}-\frac{x-89}{16}=0\)
\(\Leftrightarrow\left(x-89\right).\left(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\right)=0\)
Vì \(\frac{1}{77}+\frac{1}{78}-\frac{1}{15}-\frac{1}{16}\ne0.\)
\(\Leftrightarrow x-89=0\)
\(\Leftrightarrow x=0+89\)
\(\Leftrightarrow x=89.\)
Vậy phương trình có tập hợp nghiệm là: \(S=\left\{89\right\}.\)
Chúc bạn học tốt!
Phương trình 1:
\(\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}=10\)
\(\Rightarrow\frac{x-85}{15}+\frac{x-74}{13}+\frac{x-67}{11}+\frac{x-64}{9}-10=0\)
\(\Rightarrow\left(\frac{x-85}{15}-1\right)+\left(\frac{x-74}{13}-2\right)+\left(\frac{x-67}{11}-3\right)+\left(\frac{x-64}{9}-4\right)=0\)
\(\Rightarrow\frac{x-85-15}{15}+\frac{x-74-26}{13}+\frac{x-67-33}{11}+\frac{x-64-36}{9}=0\)
\(\Rightarrow\frac{x-100}{15}+\frac{x-100}{13}+\frac{x-100}{11}+\frac{x-100}{9}=0\)
\(\Rightarrow\left(x-100\right)\left(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\right)=0\)
Do \(\frac{1}{15}+\frac{1}{13}+\frac{1}{11}+\frac{1}{9}\ne0\)
\(\Rightarrow x-100=0\)
\(\Rightarrow x=100\)
Vậy x = 100.
Phương trình 3:
\(\frac{1909-x}{91}+\frac{1907-x}{93}+\frac{1905-x}{95}+\frac{1903-x}{97}+4=0\)
\(\Rightarrow\left(\frac{1909-x}{91}+1\right)+\left(\frac{1907-x}{93}+1\right)+\left(\frac{1905-x}{95}+1\right)+\left(\frac{1903-x}{97}+1\right)=0\)
\(\Rightarrow\frac{1909-x+91}{91}+\frac{1907-x+93}{93}+\frac{1905-x+95}{95}+\frac{1903-x+97}{97}=0\)
\(\Rightarrow\frac{2000-x}{91}+\frac{2000-x}{93}+\frac{2000-x}{95}+\frac{2000-x}{97}=0\)
\(\Rightarrow\left(2000-x\right)\left(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\right)=0\)
Do \(\frac{1}{91}+\frac{1}{93}+\frac{1}{95}+\frac{1}{97}\ne0\)
\(\Rightarrow2000-x=0\)
\(\Rightarrow x=2000\)
Vậy x = 2000.
\(a)5-\left(x-6\right)=4\left(3-2x\right)\)
\(\Leftrightarrow5-x+6=12-8x\)
\(\Leftrightarrow-x+8x=12-5-6\)
\(\Leftrightarrow7x=1\Leftrightarrow x=\frac{1}{7}\)
a) 5-(x-6)=4(3-2x)
<=>5-x-6=12-8x
<=>-x+8x=2-5-6
<=>7x=1
<=>x=1/7
a) Ta có: \(\frac{3x-2}{6}-\frac{4-3x}{18}=\frac{4-x}{9}\)
\(\Leftrightarrow\frac{3\left(3x-2\right)}{18}-\frac{4-3x}{18}-\frac{2\left(4-x\right)}{18}=0\)
\(\Leftrightarrow9x-6-4+3x-\left(8-2x\right)=0\)
\(\Leftrightarrow12x-10-8+2x=0\)
\(\Leftrightarrow10x-18=0\)
\(\Leftrightarrow10x=18\)
hay \(x=\frac{9}{5}\)
Vậy: \(x=\frac{9}{5}\)
b) Ta có: \(\frac{2+3x}{6}-x+2=\frac{x-7}{9}\)
\(\Leftrightarrow\frac{3\left(2+3x\right)}{18}-\frac{18x}{18}+\frac{36}{18}-\frac{2\left(x-7\right)}{18}=0\)
\(\Leftrightarrow6+9x-18x+36-\left(2x-14\right)=0\)
\(\Leftrightarrow42-9x-2x+14=0\)
\(\Leftrightarrow56-11x=0\)
\(\Leftrightarrow11x=56\)
hay \(x=\frac{56}{11}\)
Vậy: \(x=\frac{56}{11}\)
c) ĐKXĐ: x∉{3;-3}
Ta có: \(\frac{6-x}{x^2-9}+\frac{2}{x+3}=\frac{-5}{x-3}\)
\(\Leftrightarrow\frac{6-x}{\left(x-3\right)\left(x+3\right)}+\frac{2\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\frac{-5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow6-x+2x-6=-5x-15\)
\(\Leftrightarrow x+5x+15=0\)
\(\Leftrightarrow6x=-15\)
hay \(x=\frac{-5}{2}\)(tm)
Vậy: \(x=\frac{-5}{2}\)
d) Ta có: \(\left(5x+2\right)\left(x^2-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+2=0\\x^2-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-2\\x^2=7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{5}\\x=\pm\sqrt{7}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-2}{5};\sqrt{7};-\sqrt{7}\right\}\)
e) ĐKXĐ: x∉{4;-4}
Ta có: \(\frac{3}{x-4}+\frac{5x-2}{x^2-16}=\frac{4}{x+4}\)
\(\Leftrightarrow\frac{3\left(x+4\right)}{\left(x-4\right)\left(x+4\right)}+\frac{5x-2}{\left(x-4\right)\left(x+4\right)}-\frac{4\left(x-4\right)}{\left(x-4\right)\left(x+4\right)}=0\)
\(\Leftrightarrow3x+12+5x-2-\left(4x-16\right)=0\)
\(\Leftrightarrow8x+10-4x+16=0\)
\(\Leftrightarrow4x+26=0\)
\(\Leftrightarrow4x=-26\)
hay \(x=\frac{-13}{2}\)(tm)
Vậy: \(x=\frac{-13}{2}\)
Ta có: \(\frac{x+1}{79}+\frac{x+4}{76}=-\frac{x+7}{73}-\frac{x+9}{71}-4\)
\(\Leftrightarrow\frac{x+1}{79}+\frac{x+4}{76}+\frac{x+7}{73}+\frac{x+9}{71}+4=0\)
\(\Leftrightarrow\frac{x+1}{79}+1+\frac{x+4}{76}+1+\frac{x+7}{73}+1+\frac{x+9}{71}+1=0\)
\(\Leftrightarrow\frac{x+80}{79}+\frac{x+80}{76}+\frac{x+80}{73}+\frac{x+80}{71}=0\)
\(\Leftrightarrow\left(x+80\right)\left(\frac{1}{79}+\frac{1}{76}+\frac{1}{73}+\frac{1}{71}\right)=0\)
Vì \(\frac{1}{79}+\frac{1}{76}+\frac{1}{73}+\frac{1}{71}>0\)
nên x+80=0
hay x=-80
Vậy: x=-80
ko có kết quả sao bạn