giúp gấp từ phần b-phần g, thanks nhiều ạ <3
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20 tháng 12 2023
e: \(\left(-4156+2021\right)-\left(119+2021-4156\right)\)
\(=-4156+2021-119-2021+4156\)
\(=\left(-4156+4156\right)+\left(2021-2021\right)-119\)
=0+0-119
=-119
g: \(315\cdot75-\left(15\cdot100-315\cdot25\right)\)
\(=315\cdot75-15\cdot100+315\cdot25\)
\(=315\left(75+25\right)-15\cdot100\)
\(=315\cdot100-15\cdot100=300\cdot100=30000\)
h: \(\left(-489\right)\cdot125-\left(125\cdot11-500\cdot25\right)\)
\(=-489\cdot125-125\cdot11+500\cdot25\)
\(=125\left(-489-11\right)+500\cdot25\)
\(=125\cdot\left(-500\right)+500\cdot25\)
\(=500\left(-125+25\right)\)
\(=500\cdot\left(-100\right)=-50000\)
Bài 2:
a: \(-415-3\left(2x-1\right)^2=-490\)
=>\(3\left(2x-1\right)^2+415=490\)
=>\(3\left(2x-1\right)^2=75\)
=>\(\left(2x-1\right)^2=25\)
=>\(\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=-4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
N
2
11 tháng 7 2021
21. massive
22. openness
23. operation
24. punctuality
25. simplicity
26. inhabitable
NT
1
15 tháng 1 2024
z4:
\(\dfrac{24}{148}=\dfrac{6}{37}=\dfrac{108}{37\cdot18}\)
\(\dfrac{-14}{-36}=\dfrac{7}{18}=\dfrac{7\cdot37}{18\cdot37}=\dfrac{259}{37\cdot18}\)
mà 108<259
nên \(\dfrac{24}{148}< \dfrac{-14}{-36}\)
z5: \(\dfrac{-26}{-72}=\dfrac{26}{72}< 1\)
\(1< \dfrac{45}{20}=\dfrac{-45}{-20}\)
Do đó: \(\dfrac{-26}{-72}< \dfrac{-45}{-20}\)
z6: \(\dfrac{14}{42}=\dfrac{1}{3}=\dfrac{1\cdot4}{3\cdot4}=\dfrac{4}{12}\)
\(\dfrac{21}{28}=\dfrac{3}{4}=\dfrac{3\cdot3}{4\cdot3}=\dfrac{9}{12}\)
mà 4<9
nên \(\dfrac{14}{42}< \dfrac{21}{28}\)
z7: \(\dfrac{-14}{-56}=\dfrac{1}{4}=\dfrac{5}{20}\)
\(\dfrac{21}{35}=\dfrac{3}{5}=\dfrac{3\cdot4}{5\cdot4}=\dfrac{12}{20}\)
mà 5<12
nên \(\dfrac{-14}{-56}< \dfrac{21}{35}\)
z8: \(10A=\dfrac{10^{201}+10}{10^{201}+1}=1+\dfrac{9}{10^{201}+1}\)
\(10B=\dfrac{10^{202}+10}{10^{202}+1}=1+\dfrac{9}{10^{202}+1}\)
\(10^{201}+1< 10^{202}+1\)
=>\(\dfrac{9}{10^{201}+1}>\dfrac{9}{10^{202}+1}\)
=>\(\dfrac{9}{10^{201}+1}+1>\dfrac{9}{10^{202}+1}+1\)
=>10A>10B
=>A>B
NT
1
7 tháng 1 2024
a: \(248\left(13-197\right)+197\left(248-13\right)\)
\(=248\cdot13-248\cdot197+197\cdot248-197\cdot13\)
\(=248\cdot13-197\cdot13\)
\(=51\cdot13=663\)
b: \(146\left(295-39\right)-295\left(146-39\right)\)
\(=146\cdot295-146\cdot39-295\cdot146+295\cdot39\)
\(=295\cdot39-146\cdot39\)
\(=39\cdot149=5811\)
c: \(\left(-245\right)\left(45-948\right)-45\left(948-245\right)\)
\(=-245\cdot45+245\cdot948-45\cdot948+45\cdot245\)
\(=245\cdot948-45\cdot948\)
\(=200\cdot948=189600\)
d: \(\left(-3\right)^4-7\cdot\left(-2\right)\cdot\left(-3\right)^3+4^3-\left(-2024\right)^0\)
\(=81+14\cdot\left(-27\right)+64-1\)
\(=144-378=-234\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
11 tháng 2 2023
\(\dfrac{20\times30}{3\times6\times25}\) = \(\dfrac{4\times5\times6\times5}{3\times6\times5\times5}\) = \(\dfrac{5\times6\times5\times4}{5\times6\times5\times3}\) =\(\dfrac{4}{3}\)
\(\dfrac{15\times28}{35\times20}\) = \(\dfrac{5\times3\times7\times4}{5\times7\times4\times5}\) = \(\dfrac{5\times7\times4\times3}{5\times7\times4\times5}\) = \(\dfrac{3}{5}\)
Lời giải:
b. $-138-3(2x^2-1)=-169$
$3(2x^2-1)=-148-(-169)=21$
$2x^2-1=21:3=7$
$2x^2=7+1=8$
$x^2=4=2^2=(-2)^2\Rightarrow x=\pm 2$
c.
$4(-x+5)^2-15=21$
$4(-x+5)^2=21+15=36$
$(-x+5)^2=36:4=9=3^2=(-3)^2$
$\Rightarrow -x+5=3$ hoặc $-x+5=-3$
$\Rightarrow x=2$ hoặc $x=8$
d.
$4(-3x^2-145)=-628$
$-3x^2-145=-628:4=-157$
$3x^2+145=157$
$3x^2=157-145=12$
$x^2=12:3=4=2^2=(-2)^2$
$\Rightarrow x=\pm 2$
e.
$252-3(-x-5)^2=225$
$3(-x-5)^2=252-225=27$
$(-x-5)^2=27:3=9$
$(x+5)^2=3^2=(-3)^2$
$\Rightarrow x+5=3$ hoặc $x+5=-3$
$\Rightarrow x=-2$ hoặc $x=-8$