Tìm x :
\(x^3\) + \(\dfrac{29}{28}\) = \(\dfrac{20}{29}\)
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Bài 1:
x/-3=9/4
nên x=-9/4*3=-27/4
2x+y=-4
=>y=-4-2x=-4-2*(-27/4)=-4+27/2=27/2-8/2=19/2
a.
\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)
\(\Leftrightarrow x\left(x+1\right).\left(x-1\right)\left(x+2\right)-24=0\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)
Đặt \(a=x^2+x-1\) , ta có pt:
\(\left(a+1\right)\left(a-1\right)-24=0\)
\(\Leftrightarrow a^2-1-24=0\)
\(\Leftrightarrow a^2-25=0\)
\(\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=5\\a=-5\end{matrix}\right.\)
*Với a = 5 ta được:
\(x^2+x-1=5\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2+3x-2x-6=0\)
\(\Leftrightarrow\left(x^2+3x\right)-\left(2x+6\right)=0\)
\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)
*Với a = -5 ta được:
\(x^2+x-1=-5\)
\(\Leftrightarrow x^2+x+4=0\)
\(\Leftrightarrow x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}=0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) ( loại)
Vậy pt có tập nghiệm là: \(s=\left\{-3;2\right\}\)
c)(ĐKXĐ: x khác 30;29)
\(\Leftrightarrow\dfrac{x-29}{30}-1+\dfrac{x-30}{29}-1=\dfrac{29}{x-30}-1+\dfrac{30}{x-29}-1\)
\(\Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}=\dfrac{x-59}{30-x}+\dfrac{x-59}{29-x}\)
\(\Leftrightarrow x=59\)(tm) or \(\dfrac{1}{30}+\dfrac{1}{29}-\dfrac{1}{30-x}-\dfrac{1}{29-x}=0\)
\(\Leftrightarrow\dfrac{-x}{30\left(30-x\right)}+\dfrac{-x}{29\left(29-x\right)}=0\)
\(\Leftrightarrow x=0\)(tm) or \(\dfrac{1}{30\left(30-x\right)}+\dfrac{1}{29\left(29-x\right)}=0\)
\(\Leftrightarrow1741-59x=0\)
\(\Leftrightarrow x=\dfrac{1741}{59}\left(tm\right)\)
Vậy S={0;\(\dfrac{1741}{59}\);59}
Xét: \(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\)
\(=\dfrac{3-2-1}{6}\)
\(=0\)
\(\rightarrow C=0\)
\(\dfrac{x+1}{29}+\dfrac{x+3}{28}=\dfrac{x+5}{27}+\dfrac{x+7}{26}\)
<=>\(\dfrac{x+1}{29}+2+\dfrac{x+3}{28}+2=\dfrac{x+5}{27}+2+\dfrac{x+7}{26}+2\)
<=>\(\dfrac{x+59}{29}+\dfrac{x+59}{28}=\dfrac{x+59}{27}+\dfrac{x+59}{26}\)
<=>\(\left(x+59\right)\left(\dfrac{1}{29}+\dfrac{1}{28}-\dfrac{1}{27}-\dfrac{1}{26}\right)=0\)
vì 1/29+1/28-1/27-1/26 khác 0 =>x+59=0<=>x=-59
vậy....
\(b,1\dfrac{5}{9}x=\dfrac{28}{29}< =>\dfrac{14}{9}x=\dfrac{28}{29}=>x=\dfrac{\dfrac{28}{29}}{\dfrac{14}{9}}=\dfrac{18}{29}\)
\(c,x:\left(-\dfrac{2}{5}\right)=\dfrac{-15}{16}< =>x=\dfrac{-15}{16}.\dfrac{-2}{5}=\dfrac{3}{8}\)
b, \(\dfrac{14}{9}.x=\dfrac{28}{29}\)
\(x=\dfrac{28}{29}:\dfrac{14}{9}\)
\(\Rightarrow x=\dfrac{18}{29}\)
c, \(x=\dfrac{-15}{16}.\left(\dfrac{-2}{5}\right)\)
\(\Rightarrow x=\dfrac{3}{8}\)
\(=28\left(\dfrac{3}{14}-\dfrac{13}{21}-\dfrac{29}{42}\right)-8\\ =28\cdot\dfrac{-23}{21}-8=-\dfrac{92}{3}-8=-\dfrac{116}{3}\)
\(C=\dfrac{3}{14}:\dfrac{1}{28}-\dfrac{13}{21}:\dfrac{1}{28}-\dfrac{29}{42}:\dfrac{1}{28}-8\)
\(=28\left(\dfrac{3}{14}-\dfrac{13}{21}-\dfrac{29}{42}\right)-8\)
\(=28\left(\dfrac{9}{42}-\dfrac{26}{42}-\dfrac{29}{42}\right)-8\)
\(=28\cdot\dfrac{-46}{42}-8\)
\(=\dfrac{-116}{3}\)
a: =>15/5*34/17<x<120/15
=>6<x<8
=>x=7
b; =>20/5*58/29<x<112/11
=>8<x<112/11
=>x=9 hoặc x=10