2/3 : x =1,4-12/5
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Bài 1:
a: \(x+\dfrac{5}{4}=\dfrac{-1}{2}\)
=>\(x=-\dfrac{1}{2}-\dfrac{5}{4}=-\dfrac{2}{4}-\dfrac{5}{4}=-\dfrac{7}{4}\)
b: \(\dfrac{3}{8}-x=\dfrac{2}{3}\)
=>\(x=\dfrac{3}{8}-\dfrac{2}{3}=\dfrac{9}{24}-\dfrac{16}{24}=-\dfrac{7}{24}\)
c: \(-x+\dfrac{1}{5}=2\dfrac{2}{5}\)
=>\(\dfrac{1}{5}-x=\dfrac{12}{5}\)
=>\(x=\dfrac{1}{5}-\dfrac{12}{5}=-\dfrac{11}{5}\)
d: \(\dfrac{2}{3}x+\dfrac{7}{10}=\dfrac{3}{10}\)
=>\(x\cdot\dfrac{2}{3}=\dfrac{3}{10}-\dfrac{7}{10}=-\dfrac{4}{10}=-\dfrac{2}{5}\)
=>\(x=-\dfrac{2}{5}:\dfrac{2}{3}=-\dfrac{2}{5}\cdot\dfrac{3}{2}=-\dfrac{3}{5}\)
e: \(\dfrac{1}{2}-3x=-\dfrac{2}{5}\)
=>\(3x=\dfrac{1}{2}+\dfrac{2}{5}=\dfrac{5}{10}+\dfrac{4}{10}=\dfrac{9}{10}\)
=>\(x=\dfrac{9}{10}:3=\dfrac{3}{10}\)
f: \(\dfrac{-3}{11}-\left(x-\dfrac{2}{5}\right)=-\dfrac{14}{22}\)
=>\(\dfrac{-3}{11}-\left(x-\dfrac{2}{5}\right)=-\dfrac{7}{11}\)
=>\(x-\dfrac{2}{5}=-\dfrac{3}{11}+\dfrac{7}{11}=\dfrac{4}{11}\)
=>\(x=\dfrac{4}{11}+\dfrac{2}{5}=\dfrac{20}{55}+\dfrac{22}{55}=\dfrac{42}{55}\)
g: \(\dfrac{7}{2}-\left[\dfrac{3}{2}-\left(x+\dfrac{7}{2}\right)\right]=-\dfrac{9}{11}\)
=>\(\dfrac{7}{2}-\dfrac{3}{2}+x+\dfrac{7}{2}=-\dfrac{9}{11}\)
=>\(x+\dfrac{11}{2}=-\dfrac{9}{11}\)
=>\(x=-\dfrac{9}{11}-\dfrac{11}{2}=\dfrac{-18}{22}-\dfrac{121}{22}=\dfrac{-139}{22}\)
\(x^2-3x+9=-xy+2y\)
\(\Rightarrow x^2-3x+xy-2y=-9\)
\(\Rightarrow\left(x^2+xy\right)-\left(2x+2y\right)-x=-9\)
\(\Rightarrow x\left(x+y\right)-2\left(x+y\right)-x=-9\)
\(\Rightarrow\left(x-2\right)\left(x+y\right)-x=-9\)
\(\Rightarrow\left(x-2\right)\left(x+y\right)-\left(x-2\right)=-7\)
\(\Rightarrow\left(x-2\right)\left(x+y-1\right)=-7\) (1)
Vì x và y nguyên nên (x-2) và (x+y-1) cũng nguyên (2)
Từ (1) và (2) suy ra:
\(\left(x-2\right);\left(x+y-1\right)\inƯ\left(-7\right)=\left\{\text{±}1;\text{±}7\right\}\)
Sau đó thì bạn lập bảng và kết luận nhé!
e: \(cos\left(\dfrac{\Omega}{2}+x\right)+cos\left(2\Omega-x\right)+cos\left(3\Omega+x\right)\)
\(=-sinx+cos\left(-x\right)+cos\left(\Omega+x\right)\)
\(=-sinx+cosx-cosx=-sinx\)
f: \(2\cdot cosx-3\cdot cos\left(\Omega-x\right)+5\cdot sin\left(\dfrac{7}{2}\Omega-x\right)+cot\left(\dfrac{3}{2}\Omega-x\right)\)
\(=2cosx+3cosx+5\cdot sin\left(2\Omega+\dfrac{3}{2}\Omega-x\right)+cot\left(\dfrac{3}{2}\Omega-x\right)\)
\(=5cosx+5\cdot sin\left(\dfrac{3}{2}\Omega-x\right)+cot\left(\dfrac{3}{2}\Omega-x\right)\)
\(=5cosx+5sin\left(\Omega+\dfrac{\Omega}{2}-x\right)+cot\left(\Omega+\dfrac{\Omega}{2}-x\right)\)
\(=5cosx-5sin\left(\dfrac{\Omega}{2}-x\right)+cot\left(\dfrac{\Omega}{2}-x\right)\)
\(=5cosx-5\cdot cosx+tanx=tanx\)
g: \(2\cdot cos\left(\dfrac{\Omega}{2}-x\right)+sin\left(5\Omega-x\right)+sin\left(\dfrac{3}{2}\Omega+x\right)+cos\left(\dfrac{\Omega}{2}+x\right)\)
\(=2\cdot sinx+sin\left(\Omega-x\right)+sin\left(\Omega+\dfrac{\Omega}{2}+x\right)-sinx\)
\(=sinx+sinx-sin\left(\dfrac{\Omega}{2}+x\right)\)
\(=2sinx-cosx\)
h: \(cos\left(5\Omega-x\right)-sin\left(\dfrac{3}{2}\Omega+x\right)+tan\left(\dfrac{3}{2}\Omega-x\right)+cot\left(3\Omega-x\right)\)
\(=cos\left(\Omega-x\right)-sin\left(\Omega+\dfrac{\Omega}{2}+x\right)+tan\left(\Omega+\dfrac{\Omega}{2}-x\right)+cot\left(\Omega-x\right)\)
\(=-cosx+sin\left(\dfrac{\Omega}{2}+x\right)+tan\left(\dfrac{\Omega}{2}-x\right)-cotx\)
\(=-cosx+cosx+cotx-cotx=0\)
1; 13 x 58 x 4 + 32 x 26 x 2 + 52 x 10
= (13 x 4) x 58 + 32 x (26 x 2) + 52 x 10
= 52 x 58 + 32 x 52 + 52 x 10
= 52 x (58 + 32 + 10)
= 52 x (90 + 10)
= 52 x 100
= 5200
2; 15 x 37 x 4 + 120 x 21 + 21 x 5 x 12
= (15 x 4) x 37 + 60 x (2 x 21) + (5 x 12) x 21
= 60 x 37 + 60 x 42 + 60 x 21
= 60 x (37 + 42 + 21)
= 60 x (79 + 21)
= 60 x 100
= 6000
a: \(209-18\cdot3+578:17\)
\(=209-54+34\)
=209-20=189
b: \(29-\left[16+3\cdot\left(51-49\right)\right]\)
\(=29-16-3\cdot2\)
=13-6
=7
c: \(80-\left[130-\left(58+2\cdot3\right)\right]\)
\(=80-\left[130-58-6\right]\)
\(=80-130+58+6\)
=64-50=14
d: \(4824-4824:24+12\)
=4836-201
=4635
Gọi số có 3 chữ số đó là \(\overline{8ab}\)
Theo đề bài ta có:
\(\overline{8ab}=11\times\overline{ab}\)
\(800+\overline{ab}=11\times\overline{ab}\)
\(11\times\overline{ab}-\overline{ab}=800\)
\(10\times\overline{ab}=800\)
\(\overline{ab}=80\) hay \(\overline{8ab}=880\)
Vậy số cần tìm là 880
\(\dfrac{2}{3}:x=1,4-\dfrac{12}{5}\)
\(\dfrac{2}{3}:x=1,4-2,4\)
\(\dfrac{2}{3}:x=-1\)
\(x=\dfrac{2}{3}:\left(-1\right)\)
\(x=-\dfrac{2}{3}\)
Vậy \(x=-\dfrac{2}{3}\)
\(\dfrac{2}{3}:x=1,4-\dfrac{12}{5}\\ \dfrac{2}{3}:x=-1\\ x=-\dfrac{2}{3}\)