x + (56 : 44) = \(\dfrac{3}{4}\)
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\(\dfrac{56^3+44^3}{100}-14.56\)
\(=\dfrac{\left(56+44\right)\left(56^2-56.44+44^2\right)}{100}-14.56\)
\(=\dfrac{100.\left(56^2-56.44+44^2\right)}{100}-14.56\)
\(=\)\(56^2-44.56+44^2-14.56\)
\(=\left(56^2-14.56\right)-\left(44.56-44^2\right)\)
\(=56\left(56-14\right)-44\left(56-44\right)\)
Ta có: \(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-4}{56}+\dfrac{x-5}{55}+\dfrac{x-6}{54}\)
\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}-\dfrac{x-60}{56}-\dfrac{x-60}{55}-\dfrac{x-60}{54}=0\)
\(\Leftrightarrow x-60=0\)
hay x=60
\(\dfrac{x}{2}=\dfrac{y}{3}\text{⇒}\dfrac{x}{10}=\dfrac{y}{15}\)
\(\dfrac{y}{5}=\dfrac{z}{4}\text{⇒}\dfrac{y}{15}=\dfrac{z}{12}\)
⇒\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-21}{-3}=7\)
⇒x=70;y=105;z=84
a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)
b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)
\(\dfrac{x-1}{59}+\dfrac{x-2}{58}+\dfrac{x-3}{57}=\dfrac{x-4}{56}+\dfrac{x-5}{55}+\dfrac{x-6}{54}\)
\(\Leftrightarrow\dfrac{x-1}{59}-1+\dfrac{x-2}{58}-1+\dfrac{x-3}{57}=\dfrac{x-4}{56}-1+\dfrac{x-5}{55}-1+\dfrac{x-6}{54}-1\)
\(\Leftrightarrow\dfrac{x-60}{59}+\dfrac{x-60}{58}+\dfrac{x-60}{57}=\dfrac{x-60}{56}+\dfrac{x-60}{55}+\dfrac{x-60}{54}\)
\(\Leftrightarrow\left(x-60\right)\left(\dfrac{1}{59}+\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}-\dfrac{1}{54}\right)=0\)
\(\Leftrightarrow x-60=0\)
\(\Rightarrow x=60\)
vậy \(S=\left\{60\right\}\)
\(\dfrac{x+1}{58}+\dfrac{x+2}{57}=\dfrac{x+3}{56}+\dfrac{x+4}{55}\)
\(\Leftrightarrow\left(\dfrac{x+1}{58}+1\right)+\left(\dfrac{x+2}{57}+1\right)=\left(\dfrac{x+3}{56}+1\right)+\left(\dfrac{x+4}{55}+1\right)\)
\(\Leftrightarrow\dfrac{x+59}{58}+\dfrac{x+59}{57}-\dfrac{x+59}{56}-\dfrac{x+59}{55}=0\)
\(\Leftrightarrow\left(x+59\right)\left(\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}\right)=0\)
\(\Leftrightarrow x+59=0\)
\(\Leftrightarrow x=-59\)
\(\dfrac{x+1}{58}+\dfrac{x+2}{59}=\dfrac{x+3}{56}+\dfrac{x+4}{55}\)
\(\Leftrightarrow\dfrac{x+1}{58}+1+\dfrac{x+2}{57}+1=\dfrac{x+3}{56}+1+\dfrac{x+4}{55}+1\)
\(\Leftrightarrow\dfrac{x+59}{58}+\dfrac{x+59}{57}=\dfrac{x+59}{56}+\dfrac{x+59}{55}\)
\(\Leftrightarrow\dfrac{x+59}{58}+\dfrac{x+59}{57}-\dfrac{x+59}{56}-\dfrac{x+59}{55}=0\)
\(\Leftrightarrow\left(x+59\right)\left(\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}\right)=0\)
Mà \(\dfrac{1}{58}+\dfrac{1}{57}-\dfrac{1}{56}-\dfrac{1}{55}\ne0\)
\(\Rightarrow x+59=0\)
\(\Leftrightarrow x=-59\)
Vậy: \(S=\left\{-59\right\}\)
a, 64 x 7 + 81 x 93 + 17 x 7
= 7 x ( 64 + 17 ) + 81 x 93
= 7 x 81 + 81 x 93
= 81 x ( 93 + 7 )
= 81 x 100
= 8100
a) 64 x 7 + 81 x 93 + 17 x 7
= 7 x ( 64 + 17 ) + 81 x 93
= 7 x 81 + 81 x 93
= 81 x ( 7 + 93 )
= 81 x 100
= 8100
\(A=\left(x+\dfrac{4}{x}\right)+5\left(\dfrac{y}{3}+\dfrac{3}{y}\right)+17\left(x+y\right)\)
\(A\ge2\sqrt{\dfrac{4x}{x}}+5.2\sqrt{\dfrac{3y}{3y}}+17.5=99\)
Dấu "=" xảy ra khi \(\left(x;y\right)=\left(2;3\right)\)
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
Ta có: \(x+5^6:44=\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{3}{4}-\dfrac{15625}{44}\)
hay \(x=-\dfrac{3898}{11}\)
Vậy: \(x=-\dfrac{3898}{11}\)
\(x+\left(5^6:44\right)=\dfrac{3}{4}\)
\(x+\left(15625:44\right)=\dfrac{3}{4}\)
\(x+\dfrac{15625}{44}=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}-\dfrac{15625}{44}\)
\(x=\dfrac{-3898}{11}\)