\(\left(2x-14\right):12=26\)

\(2x-14=26.12\)

\(2x-14=312\)

\(2x=312+14\)

\(2x=326\)

\(x=326:2\)

\(x=163\)

2x-14=26*12

2x-14=312 

2x=312-14

2x=298

x= 298:2

x=149

\(3^{30}:3^{27}=3^{30-27}=3^3=27\)

Vậy x=27

27

\(\left(x:32+15\right):4=24\)

\(x:32+15=24.4\)

\(x:32+15=96\)

\(x:32=96-15\)

\(x:32=81\)

\(x=81.32\)

\(x=2592\)

Sửa bài

\(\left(x:32+15\right):4=24\)

\(x:32+15=24.4\)

\(x:32+15=6\)

\(x:32=6-15\)

\(x:32=-9\)

\(x=\left(-9\right).32\)

\(x=-288\)

\(x+2y+xy=2\)

\(\Rightarrow x\left(1+y\right)+2y+2-2=2\)

\(\Rightarrow x\left(1+y\right)+2\left(y+1\right)=4\)

\(\Rightarrow\left(y+1\right)\left(x+2\right)=4\)

\(\Rightarrow\left(x+2\right);\left(y+1\right)\in\left\{-1;1;-2;2;-4;4\right\}\)

\(\Rightarrow\left(x;y\right)\in\left\{\left(-3;-5\right);\left(-1;3\right);\left(-4;-3\right);\left(0;1\right);\left(-6;-2\right);\left(2;0\right)\right\}\)

Số viên bi của Hùng bằng nhau Đức là :

\(780:2=390\left(viên\right)\)

Số viên bi của Hùng là :

\(390+\dfrac{1}{6}x390=455\left(viên\right)\)

Số viên bi của Đức là :

\(780-455=325\left(viên\right)\)

giải nhanh giúp mik nhé 5h nộp rồi 

\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt{\dfrac{19}{16}}-\sqrt{\left(-0,75\right)^2}\\ \Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt{19}}{4}-\dfrac{3}{4}\\ \Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt{19}-3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt{19}}{4}\)

\(TH_1:x\le\dfrac{2}{3}\\ 2-3x=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{3+\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{3+\sqrt{19}}{12}\left(tm\right)\)

\(TH_2:x>\dfrac{2}{3}\\ 3x-2=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{13-\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{13-\sqrt{19}}{12}\left(tm\right)\)

Vậy \(x\in\left\{\dfrac{3+\sqrt{19}}{12};\dfrac{13-\sqrt{19}}{12}\right\}\)

\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt[]{\dfrac{19}{16}}-\sqrt[]{\left(-0,75\right)^2}\)

\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-0,75\)

\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-\dfrac{3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt[]{19}}{4}+\dfrac{3}{4}\)

\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt[]{19}}{4}\)

\(\Rightarrow\left[{}\begin{matrix}2-3x=\dfrac{5-\sqrt[]{19}}{4}\\2-3x=\dfrac{-5+\sqrt[]{19}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=2-\dfrac{5-\sqrt[]{19}}{4}\\3x=2-\dfrac{\sqrt[]{19}-5}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=\dfrac{3+\sqrt[]{19}}{4}\\3x=\dfrac{13-\sqrt[]{19}}{4}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt[]{19}}{12}\\x=\dfrac{13-\sqrt[]{19}}{12}\end{matrix}\right.\)

Chúc bạn học tốt

Xét hình thang ABCD ta có :

\(\widehat{A}+\widehat{B}+\widehat{C}+\widehat{D}=360^o\)

mà \(\left\{{}\begin{matrix}\widehat{B}+\widehat{D}=180^o\left(đề.bài\right)\\\widehat{B}+\widehat{A}=180^o\left(t/c.hình.thang\right)\end{matrix}\right.\)

\(\Rightarrow\widehat{C}=\widehat{D}\)

⇒ ABCD là hình thang cân (dpcm)

Ta có : AB // CD ⇒ \(\widehat{B}\) + \(\widehat{C}\) = 180^o mà \(\widehat{B}+\widehat{D}=\) 180^o ⇒ \(\widehat{D}=\widehat{C}\)

Vì AB // CD; \(\widehat{D}=\widehat{C}\) vậy ABCD là hình thang cân

a) \(x.125=32250\)

\(x=32250:125\)

\(x=258\)

b) \(39975:x=123\)

\(x=39975:123\)

\(x=325\)

\(x\times125=32250\)

\(\Rightarrow x=258\)

_____

\(39975:x=123\)

\(\Rightarrow x=325\)

a, 24.36 + 0,24 .36 + 24,24.64

= 36.( 24 + 0,24) + 24,24.64

= 36.24,24 + 24,24.64

= 24,24.(36 + 64)

= 24,24.100

= 2424 

b, \(5\dfrac{3}{5}\) + 3,75 + 7\(\dfrac{1}{8}\) + 2,875 + 4,4

= 5 +( 0,6 + 4,4) +  3,75 + 7 + 0,125 + 2,875

= 5 + 5 + 10,75  + 3

= 23,75