a)\(4590:\left(x-24+14\right)=30\)
b)\(61x\left(126:x\right)=1281\)
Những câu hỏi liên quan
tìm x
a, 4590 : ( x -24 + 14 ) = 30
b , 61 x ( 126 : x ) = 1281
a) \(4590\div\left(x-24+14\right)=30\)
\(\Leftrightarrow\)\(x-10=4590\div30=153\)
\(\Leftrightarrow\)\(x=153+10=163\)
Vậy....
b) \(61\times\left(126\div x\right)=1281\)
\(\Leftrightarrow\)\(126\div x=1281\div61=21\)
\(\Leftrightarrow\)\(x=126\div21=6\)
Vậy....
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Bài 2 : Tìm x
a) 4590 : ( x - 24 + 14 ) bằng 30 ; b) 61 x ( 126 : x ) bằng 1281
a) \(4590:\left(x-24+14\right)=30\)
\(x-24+14=4590:30\)
\(x-24+14=153\)
\(x-24=153-14\)
\(x-24=139\)
\(x=139+24\)
\(x=163\)
b) \(61\times\left(126:x\right)=1281\)
\(126:x=1281:61\)
\(126:x=21\)
\(x=126:21\)
\(x=6\)
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b. 61 × ( 126 : x ) = 1281
x = 126 : x = 1281 ÷ 61 = 21
x = 126÷21
x = 6
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a) 4590 ÷ ( x− 24 + 14 ) = 30
4590 ( x- 10 ) = 30
x - 10 = 4590 : 30
x-10 = 153
x = 153+10
x = 163
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61x(126:x)=1281
text{Tìm x, biết:}a) left(19x+2.5^2right):14left(13-8right)^2-4^2b) x+left(x+1right)+left(x+2right)+...+left(x+30right)1240c) 11-left(-53+xright)97d) -left(x+84right)+213-16
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\(\text{Tìm x, biết:}\)
\(a\)) \(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
\(b\)) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)
\(c\)) \(11-\left(-53+x\right)=97\)
\(d\)) \(-\left(x+84\right)+213=-16\)
Tìm x
a) 4590: (x - 24 + 14) = 30
a) \(\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
b) \(x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)
c) \(|x+7|=20+5.\left(-3\right)\)
a, (19x+2.52) : 14 = (13-8)2 - 42
(19x + 2.25) : 14 = 52 - 42
(19x + 50) : 14 = 25 - 16
(19x + 50) : 14 = 9
19 x + 50 = 9.14
19x + 50 = 126
19x = 126 - 50
19x = 76
x = 76 : 19
x = 4
vậy____
b) x + (x + 1) + (x + 2) + (x + 3)+.....+(x+30) = 1240
(x+x+x+...+x) + (1+2+3+...+30) = 1240
31x + 465 = 1240
31x = 1240 - 465
31x = 775
x = 775 : 31
x = 25
vậy____
c) |x + 7| = 20 + 5.(-3)
|x + 7| = 20 + (-15)
|x + 7| = 5
\(\Rightarrow\orbr{\begin{cases}x+7=5\\x+7=-5\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5-7\\x=-5-7\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-2\\x=-12\end{cases}}\)
vậy_____
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14 tháng 8 2023 lúc 10:52
2. Tìm x biết:
a)dfrac{2}{left(x+2right)left(x+4right)} + dfrac{4}{left(x+4right)left(x+8right)} + dfrac{6}{left(x+8right)left(x+14right)} dfrac{x}{left(x+2right)left(x+14right)}.
b)dfrac{x}{2023} + dfrac{x+1}{2022} + dfrac{x+2}{2021} +...+ dfrac{x+2022}{1} + 2023 0.
Gíup mình giải 2 bài này với!
Cảm ơn các bạn rất nhiều!!!
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2. Tìm x biết:
a)\(\dfrac{2}{\left(x+2\right)\left(x+4\right)}\) + \(\dfrac{4}{\left(x+4\right)\left(x+8\right)}\) + \(\dfrac{6}{\left(x+8\right)\left(x+14\right)}\) = \(\dfrac{x}{\left(x+2\right)\left(x+14\right)}\).
b)\(\dfrac{x}{2023}\) + \(\dfrac{x+1}{2022}\) + \(\dfrac{x+2}{2021}\) +...+ \(\dfrac{x+2022}{1}\) + 2023 = 0.
Gíup mình giải 2 bài này với!
Cảm ơn các bạn rất nhiều!!!
a/
\(VT=\dfrac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\dfrac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\dfrac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\dfrac{12}{\left(x+2\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\left(x\ne-2;x\ne-14\right)\)
\(\Rightarrow x=12\)
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\(\dfrac{x}{2023}+\dfrac{x+1}{2022}+...+\dfrac{x+2022}{1}+2023=0\)
\(\dfrac{1}{2023}x+\dfrac{1}{2022}x+\dfrac{1}{2022}\cdot1+...+\dfrac{1}{1}x+\dfrac{1}{1}\cdot2022+2023=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)+\left(\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\right)=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)=\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2022}{2022}+\dfrac{2}{2021}+\dfrac{2021}{2021}+...+\dfrac{2022}{1}+\dfrac{1}{1}}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{1}}{\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{1}}=2023\)
Vậy x = 2023
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Tìm x, biết:
a) \(\left|x-24\right|+\left|y+8\right|=1\)
b)\(\left(x-2\right)^{10}+\left|y-2\right|=0\)
c)\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+30\right)=1240\)
d)\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+2017+2018=2018\)
Giải thích cụ thể giúp mk nha
Phân tích các đa thức sau thành nhân tử ( đặt biến phụ )
a. left(x^2+xright)^2-14left(x^2+xright)+24
b. left(x^2+xright)^2+4x^2+4x-12
c. x^4+2x^3+5x^2+4x-12
d.left(x+1right)left(x+2right)left(x+3right)left(x+4right)+1
e. left(x+1right)left(x+3right)left(x+5right)left(x+7right)+15
f. left(x+1right)left(x+2right)left(x+3right)left(x+4right)-24
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Phân tích các đa thức sau thành nhân tử ( đặt biến phụ )
a. \(\left(x^2+x\right)^2-14\left(x^2+x\right)+24\)
b. \(\left(x^2+x\right)^2+4x^2+4x-12\)
c. \(x^4+2x^3+5x^2+4x-12\)
d.\(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
e. \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
f. \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
a) \(\left(x^2+x\right)^2-14\left(x^2+x\right)+24\)
Đặt \(x^2+x=y\) ta được:
\(y^2-14y+24\)
\(=x\left(y-12\right)-2\left(y-12\right)\)
\(=\left(y-2\right)\left(y-12\right)\)
Thay ngược trở lại:
\(\left(x^2+x-2\right)\left(x^2+x-12\right)\)
\(=\left(x-1\right)\left(x+2\right)\left(x-3\right)\left(x+4\right)\)
d) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+10\right)+1\)
Đặt \(x^2+5x+4=a\) được:
\(a\left(a+6\right)+1\)
\(=a^2+6a+1\)
\(=a^2+2.a.3+3^2-8\)
\(=\left(a+3\right)^2-\left(\sqrt{8}\right)^2\)
\(=\left(a+3-\sqrt{8}\right)\left(a+3+\sqrt{8}\right)\)
Mấy câu kia tương tự.
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