\(\frac{7^{x+2}+7^{x+1}+7^x}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)
\(\Rightarrow\frac{7^x.7^2+7^x.7^1+7^x}{57}=\frac{5^{2x}+5^{2x}.5+5^{2x}.5^3}{131}\)
\(\Rightarrow\frac{7^x.\left(7^2+7+1\right)}{57}=\frac{5^{2x}.\left(1+5+5^3\right)}{131}\)
\(\Rightarrow\frac{7^x.57}{57}=\frac{5^{2x}.131}{131}\)
\(\Rightarrow7^x=5^{2x}\)
Bạn tự làm phần còn lại nhé
Tìm x :
\(\frac{\left(7^{x+2}+7^{x+1}+7^x\right)^{57}}{57}=\frac{5^{2x}+5^{2x+1}+5^{2x+3}}{131}\)
Bài 3 Tìm x biết
a) \(\left(x-4\right)^2=\left(x-4\right)^4\)
b) \(\left(\dfrac{4}{5}\right)^{2x+7}=\dfrac{625}{256}\)
c) \(\dfrac{7^{x+2}+7^{x+1}+7^x}{57}=\dfrac{5^{2x}+5^{2x+1}+5^{2x+2}}{131}\)
tìm x biết :
a) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
b) \(\left(x.6\frac{2}{7}+\frac{3}{7}\right).2\frac{1}{5}-\frac{3}{7}=-2\)
c) \(x.3\frac{1}{4}+\left(-\frac{7}{6}\right).x-1\frac{2}{3}=\frac{5}{12}\)
d) \(5\frac{8}{17}:x+\left(-\frac{4}{17}\right):x+3\frac{1}{7}:17\frac{1}{3}=\frac{4}{11}\)
e) \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
bai 1 tim x,y,z
a,\(\frac{x}{10}\)=\(\frac{y}{15}\),x=\(\frac{7}{2}\)và x+2y-3z=20
b,2x=3y,49=57 và 4x-3y+5z=7
c,\(\frac{2x}{3}\)=\(\frac{3y}{4}\)=\(\frac{47}{5}\)và x+y+z=49
bài 14: Tìm x biết:
a/ \(-\frac{4}{5}-x=2,25\)
b/ \(\left(\frac{2x}{5}-1\right):\left(-5\right)=\frac{1}{7}\)
tìm x biết :
a) \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
b) \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
c) \(\left(x+\frac{1}{2}\right).\left(\frac{2}{3}-2x\right)=0\)
Tìm x biết:
25%x + x - \(\frac{1}{5}\)x = \(\frac{1}{5}\)\(\frac{2}{3}\)x - \(\frac{1}{2}\)x = \(\frac{3}{5}\)\(\frac{2}{5}\)x - \(1\frac{1}{2}\)x + x =\(\frac{1}{3}\)\(\left(x+\frac{2}{7}\right).\left(\frac{2}{3}-2x\right)=0\)tìm x:
1) \(\frac{x-4}{12}=\frac{x-6}{11}\)
2)\(\frac{3x+2}{-7}=\frac{x-4}{-2}\)
3)\(\frac{1-4x}{5}=\frac{2x-1}{-3}\)
+Tim x :
a ) \(\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
b )\(\left(x+\frac{1}{2}\right)^2=\frac{4}{25}\)
c ) \(\left(x+1\right)^{x+2}=\left(x-1\right)^{x+6}\) (x ϵ Z )
d ) \(\left(2x+3\right)^{2016}=\left(2x+3\right)^{2018}\)
e ) \(\frac{3}{x+2.x+5}+\frac{5}{x+5.x+10}+\frac{7}{x+10.x+17}=\frac{x}{x+2.x+17}\) Với \(x\notin\left\{-2;-5;-10;-17\right\}\)
