3^(x+2)+2022^0=(-4).(-7)
Bài 3: tìm x biết
a) x^+3x=0
b) (x-1)(x^+x+1)-x(x-2)(x+2)=7
c) x(x-2022)+4(2022-x)=0
giúp mình vs ạ , mình cần gấp 🌷
câu a chưa đủ đề em hấy
c, \(x\)(\(x\) - 2022) + 4.(2022 - \(x\)) = 0
(\(x\) - 2022).(\(x\) - 4) = 0
\(\left[{}\begin{matrix}x-2022=0\\x+4=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2022\\x=4\end{matrix}\right.\)
b, (\(x\) - 1)(\(x^2\) + \(x\) + 1) - \(x\)(\(x\) - 2)(\(x\) + 2) = 7
\(x^3\) - 1 - \(x\).(\(x^2\) - 4) = 7
\(x^3\) - 1 - \(x^3\) + 4\(x\) = 7
(\(x^3\) - \(x^3\)) - 1 + 4\(x\) = 7
- 1 + 4\(x\) = 7
4\(x\) = 7 + 1
4\(x\) = 8
\(x\) = 8:4
\(x\) = 2
Tính B = \(13x^7-5y^3+2022\) tại x,y thỏa mãn: \(\left|x-1\right|+\left(y+2\right)^{2022}=0\)
\(\left|x-1\right|+\left(y+2\right)^{2022}=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x-1\right|=0\\\left(y+2\right)^{2022}=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\\ \Rightarrow B=13.1-5\left(-8\right)+2022=13+40+2022=2075\)
|x-1|+(y+2)2022=0
Do |x-1| và (y+2)2022 đều ≥0⇒\(\left\{{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
⇒B=13.(1)7-5.(-2)3+2022=13+40+2022=2075
[ 155 - 15 . ( 2 . 5^2 - 3 . 4^2)] : ( 12 - 7 )^3 + 2022^0
[155 - 15.(2.52 - 3.42)] : (12 - 7)3 + 20220
= [155 - 15.(2.25 - 3.16)] : 53 + 1
= [155 - 15.(50 - 48)] : 125 + 1
= [155 - 15.2] : 125 + 1
= [155 - 30] : 125 +1
= 125 : 125 + 1
= 1 + 1
= 2
Bài1:Thực hiện phép tính:
a,(2^4.3.5^2):{450:[450-(4.5^3-2^3.5^2)]}
b,3^3.5^2-20{90-[164-2.(7^8:7^6+7^0)]}
c,[(18^7:18^6-17).2022-1986].5.1^2022-13^2.2020^0
Bài2:Tìm x:
a,(2^x+1)^2+3.(2^2+1)=2^2.10
b,3.(x-7)+2.(x+5)=41
(GIÚP MIK VỚI Ạ)
Bài 1.
\(a,\left(2^4\cdot3\cdot5^2\right):\left\{450:\left[450-\left(4\cdot5^3-2^3\cdot5^2\right)\right]\right\}\)
\(=\left(16\cdot3\cdot25\right):\left\{450:\left[450- \left(4\cdot125-8\cdot25\right)\right]\right\}\)
\(=\left(48\cdot25\right):\left\{450:\left[450-\left(500-200\right)\right]\right\}\)
\(=1200:\left[450:\left(450-300\right)\right]\)
\(=1200:\left(450:150\right)\)
\(=1200:3\)
\(=400\)
\(---\)
\(b,3^3\cdot5^2-20\left\{90-\left[164-2\cdot\left(7^8:7^6+7^0\right)\right]\right\}\)
\(=27\cdot25-20\left\{90-\left[164-2\cdot\left(7^2+1\right)\right]\right\}\)
\(=675-20\left\{90-\left[164-2\cdot\left(49+1\right)\right]\right\}\)
\(=675-20\left[90-\left(164-2\cdot50\right)\right]\)
\(=675-20\left[90-\left(164-100\right)\right]\)
\(=675-20\left(90-64\right)\)
\(=675-20\cdot26\)
\(=675-520\)
\(=155\)
\(---\)
\(c,\left[\left(18^7:18^6-17\right)\cdot2022-1986\right]\cdot5\cdot1^{2022}-13^2\cdot2020^0\)
\(=\left[\left(18-17\right)\cdot2022-1986\right]\cdot5\cdot1-169\cdot1\)
\(=\left(1\cdot2022-1986\right)\cdot5-169\)
\(=\left(2022-1986\right)\cdot5-169\)
\(=36\cdot5-169\)
\(=180-169\)
\(=11\)
Bài 2.
\(a) (2^x+1)^2+3\cdot(2^2+1)=2^2\cdot10\\\Rightarrow (2^x+1)^2+3\cdot(4+1)=4\cdot10\\\Rightarrow (2^x+1)^2+3\cdot5=40\\\Rightarrow (2^x+1)^2+15=40\\\Rightarrow (2^x+1)^2=40-15\\\Rightarrow (2^x+1)^2=25\\\Rightarrow (2^x+1)^2= (\pm 5)^2\\\Rightarrow \left[\begin{array}{} 2^x+1=5\\ 2^x+1=-5 \end{array} \right.\\ \Rightarrow \left[\begin{array}{} 2^x=4\\ 2^x=-6 (vô.lí) \end{array} \right. \\ \Rightarrow 2^x=2^2\\\Rightarrow x=2\)
Vậy \(x=2\).
\(---\)
\(b)3\cdot(x-7)+2\cdot(x+5)=41\\\Rightarrow 3\cdot x+3\cdot(-7)+2\cdot x+2\cdot5=41\\\Rightarrow 3x-21+2x+10=41\\\Rightarrow (3x+2x)+(-21+10)=41\\\Rightarrow 5x-11=41\\\Rightarrow 5x=41+11\\\Rightarrow 5x=52\\\Rightarrow x=\dfrac{52}{5}\)
Vậy \(x=\dfrac{52}{5}\).
\(Toru\)
a) Tính giá trị biểu thức: 2^3 . 3^2 + 7^16 : 7^14 -2022^0
b) Tìm x biết: 2x-9=3 . (-7)
a) $2^3\cdot3^2+7^{16}:7^{14}-2022^0$
$=8\cdot9+7^2-1$
$=72+49-1$
$=120$
b) $2x-9=3\cdot(-7)$
$\Rightarrow2x-9=-21$
$\Rightarrow2x=-21+9$
$\Rightarrow2x=-12$
$\Rightarrow x=-12:2=-6$
Tổng các x; y ; z thỏa mãn(x-1)^2022+(2y-1)^2022+|x+2y-z|^2022 = 0 là
A. 5/2 B. 7/2 C.-5/2 D.-7/2
Giúp mik nhanh với mik đang gấp lắm :<
24/-36 x 12/19 - 4/6 x 7/19 + 2022 mũ 0
\(=\dfrac{-2}{3}\cdot\dfrac{12}{19}-\dfrac{2}{3}\cdot\dfrac{7}{19}+1\)
=-2/3(12/19+7/19)+1
=1-2/3
=1/3
Tìm tất cả các số thực x thỏa mãn
a) (x − 4)^5 + (x − 8)^5 + (12 − 2x)^5 = 0.
b) (x + 2021)^7 + (x − 2022)^7 + (1 − 2x)^7 = 0.
-7/2022.503/3+7/2022.-508/3+7/3.(-2022)^0
\(=\dfrac{-7}{2022}\left(\dfrac{503}{3}+\dfrac{508}{3}\right)+\dfrac{7}{3}=\dfrac{-7}{2022}\cdot\dfrac{1011}{3}+\dfrac{7}{3}\)
\(=\dfrac{-7}{6}+\dfrac{7}{3}=\dfrac{-7}{6}+\dfrac{14}{6}=\dfrac{7}{6}\)