Tìm số tự nhiên \(x\) thoả mãn:
a) (13\(x\) – 122): 5 = 5;
b) 3\(x\)[82 - 2.(25 - 1)] = 2022.
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\(a,\left\{{}\begin{matrix}\left|x-3y\right|\ge0\\\left|y+4\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-3y=0\\y+4=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3y=-12\\y=-4\end{matrix}\right.\)
\(b,Sửa:\left|x-y-5\right|+\left(y+3\right)^2=0\\ \left\{{}\begin{matrix}\left|x-y-5\right|\ge0\\\left(y+3\right)^2\ge0\end{matrix}\right.\Rightarrow VT\ge0\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-y-5=0\\y+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+5=2\\y=-3\end{matrix}\right.\)
\(c,\left\{{}\begin{matrix}\left|x+y-1\right|\ge0\\\left(y-2\right)^4\ge0\end{matrix}\right.\Rightarrow VT\ge0\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+y-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-y=-1\\y=2\end{matrix}\right.\)
\(d,\left\{{}\begin{matrix}\left|x+3y-1\right|\ge0\\3\left|y+2\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+3y-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-3y=7\\y=-2\end{matrix}\right.\)
\(e,Sửa:\left|2021-x\right|+\left|2y-2022\right|=0\\ \left\{{}\begin{matrix}\left|2021-x\right|\ge0\\\left|2y-2022\right|\ge0\end{matrix}\right.\Rightarrow VT\ge0\)
Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}2021-x=0\\2y-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\y=1011\end{matrix}\right.\)
Bài 9,
62x73+36x33=36x73+36x27=36(73+27)=36x100=3600.
197-\([\)6x(5-1)2+20220\(]\):5=197-\([\)6x16+1\(]\):5=197-97:5=197-97/5=888/5.
Bài 10,
21-4x=13
=>4x=21-13=8
=>x=8:4=2.
30:(x-3)+1=45:43=42=16
=>30:(x-3)=16-1=15
=>x-3=30:15=2
=>x=2+3=5.
(x-1)3+5x6=38
=>(x-1)3+30=38
=>(x-1)3=38-30=8=23
=>x-1=2
=>x=3.
a: \(\left(2x-y+7\right)^{2022}>=0\forall x,y\)
\(\left|x-1\right|^{2023}>=0\forall x\)
=>\(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}>=0\forall x,y\)
mà \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}< =0\forall x,y\)
nên \(\left(2x-y+7\right)^{2022}+\left|x-1\right|^{2023}=0\)
=>\(\left\{{}\begin{matrix}2x-y+7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2x+7=9\end{matrix}\right.\)
\(P=x^{2023}+\left(y-10\right)^{2023}\)
\(=1^{2023}+\left(9-10\right)^{2023}\)
=1-1
=0
c: \(\left|x-3\right|>=0\forall x\)
=>\(\left|x-3\right|+2>=2\forall x\)
=>\(\left(\left|x-3\right|+2\right)^2>=4\forall x\)
mà \(\left|y+3\right|>=0\forall y\)
nên \(\left(\left|x-3\right|+2\right)^2+\left|y+3\right|>=4\forall x,y\)
=>\(P=\left(\left|x-3\right|+2\right)^2+\left|y-3\right|+2019>=4+2019=2023\forall x,y\)
Dấu '=' xảy ra khi x-3=0 và y-3=0
=>x=3 và y=3
\(a,2^x+2^{x+3}=144\\ 2^x.\left(1+2^3\right)=144\\ 2^x.9=144\\ 2^x=144:9\\ 2^x=16=2^4\\ vậy:x=4\)
\(b,\left(x-5\right)^{2022}=\left(x-5\right)^{2021}\\ Vì:\left[{}\begin{matrix}0^{2022}=0^{2021}\\1^{2022}=1^{2021}\end{matrix}\right.\\ Vậy:\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
a) 7 + x = 362 => x = 362 - 7 => x = 355
Vậy x = 355
b) 25 - x = 15 => x = 25 – 15 => x = 10
Vậy x = 10
c) x - 56 = 4 => x = 56 + 4 => x = 60.
Vậy x = 60
1) (x - 35) - 120 = 0
x - 35 = 120
x = 120 + 35
x = 155
2) 310 - (118 - x) = 217
118 - x = 310 - 217
118 - x = 93
x = 118 - 93
x = 25
3) 156 - (x + 61) = 82
x + 61 = 156 - 82
x + 61 = 74
x = 74 - 61
x = 13
4) 814 - (x - 305) = 712
x - 305 = 814 - 712
x - 305 = 102
x = 102 + 305 = 407
5) 100 - 7 - (x - 5) = 58
x - 5 = 93 - 58
x - 5 = 35
x = 35 + 5 = 40
6) 12(x - 1) : 3 = 43 + 23
4(x - 1) = 72
x - 1 = 18
x = 18 + 1 = 19
7) 24 + 5x = 75 : 73
24 + 5x = 49
5x = 25
x = 25 : 5 = 5
8) 5(x - 1) : 3 = 43 + 23
\(\dfrac{5}{3}\left(x-1\right)=72\)
x - 1 = \(\dfrac{216}{5}\)
x = 221/5
9) 5(x - 4)2 - 7 = 13
5(x - 4)2 = 20
(x - 4)2 = 4
\(\Rightarrow\left[{}\begin{matrix}x-4=2\\x-4=-2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\end{matrix}\right.\)
10) (x + 1) + (x + 2) + ... + (x + 30) = 795
=> (x + x + x + ... + x) + (1 + 2 + 3 +...+ 30) = 795 (1)
Đặt A = 1 + 2 + 3 +...+ 30
Số số hạng trong A là: (30 - 1) : 1 + 1 = 30 (số)
Tổng A bằng : (30 + 1).30 : 2 =465
Thay A = 465 vào (1) , ta được:
30x + 465 = 795
=> 30x =330
=> x =11
1: =>x-35=120
=>x=120+35=155
2: =>118-x=310-217=93
=>x=118-93=25
3: =>x+61=156-82=74
=>x=74-61=13
4: =>x-305=814-712=102
=>x=102+305=407
5: =>93-(x-5)=58
=>x-5=35
=>x=40
6: =>4(x-1)=64+8=72
=>x-1=18
=>x=19
7: =>5x+24=49
=>5x=25
=>x=5
8: =>5(x-1):3=4^3+2^3=64+8=72
=>5(x-1)=216
=>x-1=216/5
=>x=221/5
\(3x-2^4=5^3\\ 3x-16=125\\ 3x=125+16=141\\ x=\dfrac{141}{3}=47\)
bài 7:a thực hiện phép tính .81 x 2022 + 25 x 2022 - 6 x 2022 .B Tìm x biết ( x - 1 ) 2/3 - 1/5= 2/5
\(a,81\cdot2022+25\cdot2022-6\cdot2022=2022\cdot\left(81+25-6\right)=2022\cdot100=202200\)
\(b,\left(x-1\right)\cdot\frac{2}{3}-\frac{1}{5}=\frac{2}{5}\)
\(\left(x-1\right)\cdot\frac{2}{3}=\frac{3}{5}\)
\(x-1=\frac{9}{10}\)
\(x=\frac{19}{10}\)
Vậy \(x=\frac{19}{10}\)
( Nếu phần b là hỗn số thì mình làm thế kia , còn nếu là nhân thì bạn tham khảo Câu hỏi của lương bảo ngọc - Toán lớp 5 - Học trực tuyến OLM nhé )
81 x 2022 + 25 x 2022 - 6 x 2022
= ( 81 + 25 - 6 ) x 2022
= 100 x 2022
= 202 200
b) \(\left(\text{x - 1}\right)\frac{\text{2}}{\text{3}}-\frac{\text{1}}{\text{5}}=\frac{\text{2}}{\text{5}}\)
\(\frac{\text{3 x }\text{( x - 1 ) }+\text{2}}{\text{3}}=\frac{\text{1}}{\text{5}}+\frac{\text{2}}{\text{5}}=\frac{\text{3}}{\text{5}}\)
=> \(\text{3 x ( x - 1 ) }+\text{2}=\frac{\text{3}}{\text{5}}\text{ x 3 = }\frac{\text{9}}{\text{5}}\)
=> \(\text{3 x ( x - 1 ) }=\frac{\text{9}}{\text{5}}-\text{2}=\frac{\text{-1}}{\text{5}}\)
=> \(\text{ x-1}=\frac{\text{-1}}{\text{5}}:3=\frac{\text{-1}}{\text{15}}\)
=> \(\text{x}=\frac{\text{-1}}{\text{15}}+\text{1 = }\frac{\text{14}}{\text{15}}\)
Bài 1:vì 15 chia hết cho 5 suy ra 2022.15 chia hết cho 5
vì 25 chia hết cho 5 suy ra 2022.15 + 25 chia hết cho 5
a)
\(\begin{array}{l}\left( {13x{\rm{ }}-{\rm{ }}{{12}^2}} \right):{\rm{ }}5{\rm{ }} = {\rm{ }}5\\13x{\rm{ }}-{\rm{ }}{12^2} = 5.5\\13x{\rm{ }}-{\rm{ }}144 = 25\\13x = 25 + 144\\13x = 169\\x = 13\end{array}\)
Vậy \(x = 13\)
b)
\(\begin{array}{l}3x\left[ {{8^2} - 2.\left( {{2^5} - {\rm{ }}1} \right)} \right]{\rm{ }} = {\rm{ }}2022\\3x\left[ {64 - 2.\left( {32 - {\rm{ }}1} \right)} \right]{\rm{ }} = {\rm{ }}2022\\3x\left[ {64 - 2.31} \right]{\rm{ }} = {\rm{ }}2022\\3x\left( {64 - 62} \right){\rm{ }} = {\rm{ }}2022\\3x.2 = 2022\\6x = 2022\\x = 337\end{array}\)
Vậy \(x = 337.\)