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\(\left(-2023\right).2024-2023.\left(10-2024\right)\\ =-2023.2024-2023.10+2023.2024\\ =\left(2023.2024-2023.2024\right)-2023.10\\ =0-2023.10=-20230\\ ---\\ 198.246-246.\left(198-19\right)\\ =198.246-246.198+246.19\\ =0-246.19=-4674\\ ---\\ 297.\left(435-24\right)-435.\left(297-24\right)\\ =297.435-297.24-435.297+435.24\\ =\left(297.435-435.297\right)+\left(435.24-297.24\right)\\ =0+\left(435-297\right).24\\ =0+138.24=3312\\ ----\\ \left(-367\right).\left(19-198\right)+198.\left(19-367\right)\\ =-367.19+367.198+198.19-367.198\\ =\left(367.198-367.198\right)+\left(198.19-367.19\right)\\ =0+19.\left(198-367\right)=0+19.\left(-169\right)=-3211\)
3:
\(6a+11b-6\left(a+7b\right)\)
\(=6a+11b-6a-42b=-31b⋮31\)
Ta có: \(\left(6a+11b\right)-6\left(a+7b\right)⋮31\)
\(6a+11b⋮31\)
Do đó: \(6\left(a+7b\right)⋮31\)
=>\(a+7b⋮31\)
Ta có: \(\left(6a+11b\right)-6\left(a+7b\right)⋮31\)
\(a+7b⋮31\)
Do đó: \(6a+11b⋮31\)
4:
\(5a+2b⋮17\)
=>\(12\left(5a+2b\right)⋮17\)
=>\(60a+24b⋮17\)
=>\(51a+17b+9a+7b⋮17\)
=>\(17\left(3a+b\right)+\left(9a+7b\right)⋮17\)
mà \(17\left(3a+b\right)⋮17\)
nên \(9a+7b⋮17\)
h: \(\left(-145\right)-5\left(2x-1\right)^3=480\)
=>\(5\left(2x-1\right)^3=-145-480\)=-625
=>(2x-1)3=-125
=>2x-1=-5
=>2x=-5+1=-4
=>x=-4/2=-2
k: \(3\left(-4+x\right)^3-\left(-4\right)^2=-97\)
=>\(3\left(x-4\right)^3-16=-97\)
=>\(3\cdot\left(x-4\right)^3=-97+16=-81\)
=>\(\left(x-4\right)^3=-27\)
=>x-4=-3
=>x=-3+4=1
n: \(195+3\left(-2x+5\right)^3=192\)
=>\(3\left(-2x+5\right)^3=192-195=-3\)
=>\(\left(-2x+5\right)^3=-1\)
=>-2x+5=-1
=>-2x=-1-5=-6
=>\(x=\dfrac{-6}{-2}=3\)
o: \(\left(-148\right)-5\left(-x+6\right)^4=-228\)
=>\(148+5\left(x-6\right)^4=228\)
=>\(5\left(x-6\right)^4=80\)
=>\(\left(x-6\right)^4=16\)
=>\(\left[{}\begin{matrix}x-6=2\\x-6=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=4\end{matrix}\right.\)
p: \(-43-\left(-4+x\right)^4=-44\)
=>\(\left(x-4\right)^4+43=44\)
=>\(\left(x-4\right)^4=1\)
=>\(\left[{}\begin{matrix}x-4=1\\x-4=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=3\end{matrix}\right.\)
q: \(3\left(-x+5\right)^2-\left(-5\right)^2\cdot\left(-2\right)^3=203\)
=>\(3\left(x-5\right)^2-25\cdot\left(-8\right)=203\)
=>\(3\left(x-5\right)^2+200=203\)
=>\(3\left(x-5\right)^2=3\)
=>(x-5)2=1
=>\(\left[{}\begin{matrix}x-5=1\\x-5=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=4\end{matrix}\right.\)
h.
$(-145)-5(2x-1)^3=480$
$5(2x-1)^3=-145-480=-625$
$(2x-1)^3=-625:5=-125=(-5)^3$
$\Rightarrow 2x-1=-5$
$\Rightarrow x=-3$
k.
$3(-4+x)^3-(-4)^2=-97$
$3(-4+x)^3-16=-97$
$3(-4+x)^3=-97+16=-81$
$(-4+x)^3=-81:3=-27=(-3)^3$
$\Rightarrow -4+x=-3$
$\Rightarrow x=-3-(-4)=1$
\(\left(1+\dfrac{1}{1}\right).\left(1+\dfrac{1}{2}\right).\left(1+\dfrac{1}{3}\right).\left(1+\dfrac{1}{4}\right).\left(1+\dfrac{1}{5}\right).\left(1+\dfrac{1}{6}\right)\)
=\(\dfrac{2}{1}.\dfrac{3}{2}.\dfrac{4}{3}.\dfrac{5}{4}.\dfrac{6}{5}.\dfrac{7}{6}=7\)
- Chọn câu D :)
316 - ( 5\(^2\) . 2\(^2\) + 2\(^4\) ) : 2\(^3\) - 3. 2\(^3\)
= 316 - ( 25 . 4 + 16 ) : 8 - 3. 8
= 316 - ( 100 + 16 ) : 8 - 3. 8
= 316 - 116 : 8 - 3. 8
= 200 : 8 - 3. 8
= 25 - 24
= 1
316 - ( 52 . 22 + 24 ) : 23 - 3. 23
= 316 - ( 25 . 4 + 16 ) : 8 - 3. 8
= 316 - ( 100 + 16 ) : 8 - 3. 8
= 316 - 116 : 8 - 3. 8
= 200 : 8 - 3. 8
= 25 - 24
= 1
\(5\left(x+60\right)=980-160\)
\(5\left(x+60\right)=820\)
\(x+60=820:5\)
\(x+60=164\)
\(x=104\)
a) \(-\dfrac{3}{20}< -\dfrac{2}{15}< \dfrac{1}{2}< \dfrac{3}{4}< \dfrac{4}{5}\)
b) \(\dfrac{13}{2};-\dfrac{8}{7};-15;\dfrac{5}{3}\) (mik rút gọn p/s nghịch đảo luôn nha)
ok hơi dài mà chắc mất thời gian lắm
1) (-5)+(-18)-(+9)+(-3)
= -23 - 6
= -29
2) (-4)-(-8)+(-15)+(-10)
= 4+ -25
=-21