tìm x(2x-14):12=26 cíu tui ạ
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\(\left(x:32+15\right):4=24\)
\(x:32+15=24.4\)
\(x:32+15=96\)
\(x:32=96-15\)
\(x:32=81\)
\(x=81.32\)
\(x=2592\)
Sửa bài
\(\left(x:32+15\right):4=24\)
\(x:32+15=24.4\)
\(x:32+15=6\)
\(x:32=6-15\)
\(x:32=-9\)
\(x=\left(-9\right).32\)
\(x=-288\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x+2y+xy=2\)
\(\Rightarrow x\left(1+y\right)+2y+2-2=2\)
\(\Rightarrow x\left(1+y\right)+2\left(y+1\right)=4\)
\(\Rightarrow\left(y+1\right)\left(x+2\right)=4\)
\(\Rightarrow\left(x+2\right);\left(y+1\right)\in\left\{-1;1;-2;2;-4;4\right\}\)
\(\Rightarrow\left(x;y\right)\in\left\{\left(-3;-5\right);\left(-1;3\right);\left(-4;-3\right);\left(0;1\right);\left(-6;-2\right);\left(2;0\right)\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 :
\(780-455=325\left(viên\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt{\dfrac{19}{16}}-\sqrt{\left(-0,75\right)^2}\\ \Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt{19}}{4}-\dfrac{3}{4}\\ \Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt{19}-3}{4}\)
\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt{19}}{4}\)
\(TH_1:x\le\dfrac{2}{3}\\ 2-3x=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{3+\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{3+\sqrt{19}}{12}\left(tm\right)\)
\(TH_2:x>\dfrac{2}{3}\\ 3x-2=\dfrac{5-\sqrt{19}}{4}\\ \Rightarrow3x=\dfrac{13-\sqrt{19}}{4}\\ \Rightarrow x=\dfrac{13-\sqrt{19}}{12}\left(tm\right)\)
Vậy \(x\in\left\{\dfrac{3+\sqrt{19}}{12};\dfrac{13-\sqrt{19}}{12}\right\}\)
\(\dfrac{1}{2}-\left|2-3x\right|=\sqrt[]{\dfrac{19}{16}}-\sqrt[]{\left(-0,75\right)^2}\)
\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-0,75\)
\(\Rightarrow\dfrac{1}{2}-\left|2-3x\right|=\dfrac{\sqrt[]{19}}{4}-\dfrac{3}{4}\)
\(\Rightarrow\left|2-3x\right|=\dfrac{1}{2}-\dfrac{\sqrt[]{19}}{4}+\dfrac{3}{4}\)
\(\Rightarrow\left|2-3x\right|=\dfrac{5-\sqrt[]{19}}{4}\)
\(\Rightarrow\left[{}\begin{matrix}2-3x=\dfrac{5-\sqrt[]{19}}{4}\\2-3x=\dfrac{-5+\sqrt[]{19}}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=2-\dfrac{5-\sqrt[]{19}}{4}\\3x=2-\dfrac{\sqrt[]{19}-5}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=\dfrac{3+\sqrt[]{19}}{4}\\3x=\dfrac{13-\sqrt[]{19}}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+\sqrt[]{19}}{12}\\x=\dfrac{13-\sqrt[]{19}}{12}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 ⇒ \(\widehat{B}\) + \(\widehat{C}\) = 180o mà \(\widehat{B}+\widehat{D}=\) 180o ⇒ \(\widehat{D}=\widehat{C}\)
Vì AB // CD; \(\widehat{D}=\widehat{C}\) vậy ABCD là hình thang cân
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x.125=32250\)
\(x=32250:125\)
\(x=258\)
b) \(39975:x=123\)
\(x=39975:123\)
\(x=325\)
\(x\times125=32250\)
\(\Rightarrow x=258\)
_____
\(39975:x=123\)
\(\Rightarrow x=325\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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
\(\left(2x-14\right):12=26\)
\(2x-14=26.12\)
\(2x-14=312\)
\(2x=312+14\)
\(2x=326\)
\(x=326:2\)
\(x=163\)
2x-14=26*12
2x-14=312
2x=312-14
2x=298
x= 298:2
x=149