\(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)
Giải phương trình trên , trình bày rõ ràng !
\(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)
Giải phương trình trên , trình bày rõ ràng !
\(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)
\(\Rightarrow\frac{x-17}{33}-1+\frac{x-21}{29}-1+\frac{x}{25}-2=0\)
\(\Rightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)
\(\Rightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)
Dễ thấy\(\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)>0\Rightarrow x-50=0\Rightarrow x=50\)
Vậy x = 50
Ta có
\(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\)
\(\Leftrightarrow\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)
\(\Leftrightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)
\(\Leftrightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)
Mà : \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\ne0\)
\(\Rightarrow x-50=0\)
\(\Rightarrow x=50\)
Vậy : \(x=50\)
\(\Leftrightarrow\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}-4=0\)
\(\Rightarrow\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)
\(\Rightarrow\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)
\(\Rightarrow\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)
Mà \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}>0\)nên
\(\Rightarrow x-50=0\Rightarrow x=50\)
Vậy x=50
Các bạn giải cụ thể từng bước dùm mình cho mình hiểu nha. C.ơn các bạn nhiều
\(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)
<=>\(\frac{x+43}{57}+1+\frac{x+46}{54}+1=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)
<=>\(\frac{x+100}{57}+\frac{x+100}{54}=\frac{x+100}{51}+\frac{x+100}{48}\)
<=>\(\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)
<=>\(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)
Vì \(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\ne0\)
=>x+100=0
<=>x=-100
k nha bạn
\(\Leftrightarrow\frac{37x+1648}{1026}=\frac{11x+556}{272}\Rightarrow\left(37x+1648\right)272=1026\left(11x+556\right)\)
<=>(37x+1648)272=272(37x+1648)
=>272(37x+1648)=1026(11x+556)
=>10064x+448256=11286x+570456
<=>-1222x=122200
=>x=122200:-1222
=>x=-100 ( dễ hiểu chưa hả )
\(\frac{x+43}{57}+\frac{x+46}{54}=\frac{x+49}{51}+\frac{x+52}{48}\)
\(\Leftrightarrow\)\(\frac{x+43}{57}+1+\frac{x+46}{54}=\frac{x+49}{51}+1+\frac{x+52}{48}+1\)
\(\Leftrightarrow\)\(\frac{x+43}{57}+\frac{57}{57}+\frac{x+46}{54}+\frac{54}{54}=\frac{x+49}{51}+\frac{51}{51}+\frac{x+52}{48}+\frac{48}{48}\)
\(\Leftrightarrow\)\(\frac{x+100}{57}+\frac{x+100}{54}-\frac{x+100}{51}-\frac{x+100}{48}=0\)
\(\Leftrightarrow\)\(\left(x+100\right)\left(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\right)=0\)
Vì \(\frac{1}{57}+\frac{1}{54}-\frac{1}{51}-\frac{1}{48}\ne0\)nên \(x+100=0\Leftrightarrow x=-100\)
Vậy \(x=-100\)
tìm x,y biết \(\frac{x}{8}=\frac{1}{y}\)mong mọi người trình bày rõ ràng giùm e
x/8=1/y<=>xy=1.8=8(nhân chéo)
=>x,y E Ư(8)
do đó
+)x=1=>y=8
+)x=8=>y=1
+)x=2=>y=4
+)x=4=>y=2
+)x=-1=>y=-8
+)x=-8=>y=-1
+)x=-2=>y=-4
+)x=-4=>y=-2
vậy ....
tìm x
x+43/57 + x+46/54 = x+49/51 + x+52/48
Giải phương trình: \(\frac{x^2}{3}+\frac{48}{x^2}=10(\frac{x}{3}-\frac{4}{x})\)
ĐKXĐ: ...
Đặt \(\frac{x}{3}-\frac{4}{x}=a\Rightarrow\frac{x^2}{9}+\frac{16}{x^2}-\frac{8}{3}=a^2\Rightarrow\frac{x^2}{9}+\frac{16}{x^2}=a^2+\frac{8}{3}\)
\(\Rightarrow\frac{x^2}{3}+\frac{48}{x^2}=3a^2+8\)
\(3a^2+8=10a\Leftrightarrow3a^2-10a+8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=\frac{4}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{x}{3}-\frac{4}{x}=2\\\frac{x}{3}-\frac{4}{x}=\frac{4}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x^2-6x-12=0\\x^2-4x-12=0\end{matrix}\right.\)
Giải phương trình :
\(\dfrac{43-x}{57}+\dfrac{46-x}{54}=\dfrac{49-x}{51}+\dfrac{52-x}{48}\)
\(\frac{43-x}{57}+\frac{46-x}{54}=\frac{49-x}{51}+\frac{52-x}{48}\)
\(\Leftrightarrow\left(\frac{43-x}{57}+1\right)+\left(\frac{46-x}{54}+1\right)=\left(\frac{49-x}{51}+1\right)+\left(\frac{52-x}{48}+1\right)\)
\(\Leftrightarrow\frac{43-x+57}{57}+\frac{46-x+54}{54}=\frac{49-x+51}{51}+\frac{52-x+48}{48}\)
\(\Leftrightarrow\frac{100-x}{57}+\frac{100-x}{54}=\frac{100-x}{51}+\frac{100-x}{48}\)
\(\Leftrightarrow\frac{100-x}{57}+\frac{100-x}{54}-\left(\frac{100-x}{51}+\frac{100-x}{48}\right)=0\)
\(\Leftrightarrow\left(100-x\right)\left[\left(\frac{1}{57}+\frac{1}{54}\right)-\left(\frac{1}{51}+\frac{1}{48}\right)\right]=0\) (*)
Vì\(\frac{1}{57}< \frac{1}{51},\frac{1}{54}< \frac{1}{48}\Rightarrow\left(\frac{1}{57}+\frac{1}{54}\right)< \left(\frac{1}{51}+\frac{1}{48}\right)\)
\(\Rightarrow\left(\frac{1}{57}+\frac{1}{54}\right)-\left(\frac{1}{51}+\frac{1}{48}\right)< 0\)
Phương trình (*) xảy ra khi: \(100-x=0\Leftrightarrow x=100\)
Vậy phương trình có nghiệm duy nhất là x = 100
Giải phương trình: \(\frac{x^2}{4}+\frac{900}{x^2}=2+48(\frac{10}{x}-\frac{x}{6})\)
ĐKXĐ: ...
Đặt \(\frac{10}{x}-\frac{x}{6}=a\Rightarrow a^2=\frac{100}{x^2}+\frac{x^2}{36}-\frac{10}{3}\Rightarrow\frac{100}{x^2}+\frac{x^2}{36}=a^2+\frac{10}{3}\)
\(\Rightarrow\frac{900}{x^2}+\frac{x^2}{4}=9a^2+30\)
Phương trình trở thành:
\(9a^2+30=2+48a\)
\(\Leftrightarrow9a^2-48a+28=0\Rightarrow\left[{}\begin{matrix}a=\frac{14}{3}\\a=\frac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{10}{x}-\frac{x}{6}=\frac{14}{3}\\\frac{10}{x}-\frac{x}{6}=\frac{2}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\frac{x^2}{6}+\frac{14}{3}x-10=0\\\frac{x^2}{6}+\frac{2}{3}x-10=0\end{matrix}\right.\)
1. Tính:
9/7 x 14/3 + 5
( Trình bày rõ ràng từng cách giải )
\(\frac{9}{7}x\frac{14}{3}+5\)
=\(\frac{9}{7}x\frac{14}{3}=6+5\)
=\(6+5=11\)
k mình nha
\(\frac{9}{7}\) \(\times\) \(\frac{14}{3}\) + 5
= \(\frac{9\times14}{7\times3}\) + 5
= 6 + 5
= 11
Giải phương trình: \(\frac{25}{x^2}-\frac{49}{(x-7)^2}=1\)
ĐKXĐ: ...
\(\Leftrightarrow\frac{49}{\left(x-7\right)^2}+1=\frac{25}{x^2}\)
\(\Leftrightarrow\frac{49x^2}{\left(x-7\right)^2}+x^2=25\)
\(\Leftrightarrow\frac{49x^2}{\left(x-7\right)^2}+2.\frac{7x}{x-7}.x+x^2-\frac{14x^2}{x-7}=25\)
\(\Leftrightarrow\left(\frac{7x}{x-7}+x\right)^2-\frac{14x^2}{x-7}=25\)
\(\Leftrightarrow\left(\frac{x^2}{x-7}\right)^2-\frac{14x^2}{x-7}-25=0\)
Đặt \(\frac{x^2}{x-7}=a\)
\(\Rightarrow a^2-14a-25=0\)
Nghiệm xấu, bạn tự giải tiếp đoạn cuối