tìm x biết:1+2+3+4+...+x=342
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quên làm bài 2
| x - | 6 - 8 | | = 16
| x - 2 | = 16
\(\Rightarrow\orbr{\begin{cases}x-2=16\\x-2=-16\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=16+2\\x=-16+2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=18\\x=-14\end{cases}}\)
a) 342 - ( |27 - 36| + 342)
= 342 - ( | -9 | + 342 )
= 342 - ( 9 + 342 )
= 342 - 9 - 342
= -9
b) |5 - |7 - ( 40 - |-421|)||
= | 5 - | 7 - ( 40 - 421 ) ||
= | 5 - | 7 + 381 | |
= | 5 - 388 |
= 383
c) | ( 7 - 5 )2 . | -23 | |
= | 22 . 8 |
= 32
d)| -16| - ( 4 - |18 - 28| )
16 - ( 4 - 10 )
= 16 - 4 + 10
= 22
1.
\(PTK_{CuSO_4}=64+32+16.4=160\left(đvC\right)\)
\(PTK_{5CaCO_3}=5\left(40+12+16.3\right)=500\left(đvC\right)\)
\(PTK_{Ca\left(OH\right)_2}=40+\left(16+1\right).2=74\left(đvC\right)\)
2.
Theo đề, ta có:
\(d_{\dfrac{X}{Mg}}=\dfrac{M_X}{M_{Mg}}=\dfrac{M_X}{24}=\dfrac{4}{3}\left(lần\right)\)
=> MX = 32(g)
Vậy X là lưu huỳnh (S)
3.
Ta có: \(PTK_{Al_x\left(SO_4\right)_3}=27.x+\left(32+16.4\right).3=342\left(đvC\right)\)
=> x = 2
Bài 1.Phân tử khối các chất:
\(CuSO_4\)\(\Rightarrow64+32+4\cdot16=160\left(đvC\right)\)
\(CaCO_3\Rightarrow40+12+3\cdot16=100\left(đvC\right)\)
\(Ca\left(OH\right)_2\Rightarrow40+16\cdot2+2=74\left(đvC\right)\)
Bài 2.Theo bài: \(\overline{M_X}=\dfrac{4}{3}\overline{M_{Mg}}=\dfrac{4}{3}\cdot24=32\left(đvC\right)\)
Vậy X là lưu huỳnh.KHHH: S.
Bài 3. \(Al_x\left(SO_4\right)_3\) \(\Rightarrow27x+3\cdot\left(32+4\cdot16\right)=342\Leftrightarrow x=2\)
4^x+342=7^y
4^x phải lẻ vì 7^y lúc nào cũng lẻ
=> x =0 ( 4^0 = 1 ; 1 lẻ )
có 7^y=342+1
=> 7^y = 343
=> 7^3=343
=> y =3
K nhé
\(x\left(x+1\right)=156\)
\(\Rightarrow x^2+x=156\)
\(\Rightarrow x^2+x-156=0\)
\(\Rightarrow x^2+13x-12x-156=0\)
\(\Rightarrow x\left(x+13\right)-12\left(x+13\right)=0\)
\(\Rightarrow\left(x+13\right)\left(x-12\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=12\\x=-13\end{matrix}\right.\)
___________________
\(x\left(x+1\right)=342\)
\(\Rightarrow x^2+x=342\)
\(\Rightarrow x^2+x-342=0\)
\(\Rightarrow x^2+19x-18x-342=0\)
\(\Rightarrow x\left(x+19\right)-18\left(x+19\right)=0\)
\(\Rightarrow\left(x+19\right)\left(x-18\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-19\\x=18\end{matrix}\right.\)
__________________
\(x\left(x+1\right)=650\)
\(\Rightarrow x^2+x=650\)
\(\Rightarrow x^2-x+650=0\)
\(\Rightarrow x^2+26x-25x-650=0\)
\(\Rightarrow x\left(x+26\right)-25\left(x+26\right)=0\)
\(\Rightarrow\left(x+26\right)\left(x-25\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-26\\x=25\end{matrix}\right.\)
______________________
\(x\left(x+1\right)=380\)
\(\Rightarrow x^2+x=380\)
\(\Rightarrow x^2+x-380=0\)
\(\Rightarrow x^2+20x-19x-380=0\)
\(\Rightarrow x\left(x+20\right)-19\left(x+20\right)=0\)
\(\Rightarrow\left(x+20\right)\left(x-19\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-20\\x=19\end{matrix}\right.\)
a, \(x\).(\(x\) + 1) = 156
156 = 22.3.13 = 12.13
Vậy \(x\).(\(x\) + 1) = 12.13
Vậy \(x\) = 12
b, \(x.\)(\(x\) + 1) = 342
342 = 2.32.19 = 18.19
\(x\).(\(x+1\)) = 18.19
\(x\) = 18
c, \(x\).(\(x\) + 1) = 650
650 = 2.52.13 = 25.26
\(x\).(\(x\) +1) = 25.26
\(x\) = 25
d, \(x\).(\(x\) +1) = 380
380 = 22.5.19 = 19.20
\(x\).(\(x\) + 1) = 19.20
\(x\) = 19
X=1 vì ...=331