cho M= 7+72+73+....+72021. Tìm x để 6M +7 = 7x
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\(A=\left(1+7\right)+...+7^{2020}\left(1+7\right)=8\left(1+...+7^{2020}\right)⋮8\)
\(A = (1 + 7) +...+7^2\)\(^0\)\(^2\)\(^0\) \((1 + 7) = 8 (1+...+7^2\)\(^0\)\(^2\)\(^0\)\() \) ⋮\(8\)
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a/ (d3)//(d4) \(\Leftrightarrow\left\{{}\begin{matrix}m^2+6m=7\\2n+7\ne-n^2-9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(m-1\right)\left(m+7\right)=0\\n^2+2n+16\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}m=1\\m=-7\end{matrix}\right.\\\left(n+1\right)^2+16\ne0\forall n\in R\end{matrix}\right.\)
KL: m=1 hoặc m=-7, \(n\in R\) thỏa mãn đề bài
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c) \(\left|x\right|=3,5\Rightarrow\left[{}\begin{matrix}x=3,5\\x=-3,5\end{matrix}\right.\)
d) \(\left|x\right|=-2,7\Rightarrow x\in\varnothing\)
l) \(\left|x+\dfrac{3}{4}\right|-5=-2\Rightarrow\left|x+\dfrac{3}{4}\right|=3\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=3\\x+\dfrac{3}{4}=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3-\dfrac{3}{4}\\x=-3-\dfrac{3}{4}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{9}{4}\\x=\dfrac{15}{4}\end{matrix}\right.\)
Đính chính câu l \(x=-\dfrac{15}{4}\) không phải \(x=\dfrac{15}{4}\)
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M = 7 + 72 + 73 + 74 + ..... + 7100
M = 7+(1+7)+73+(1+7)+...+799+(1+7)
M = 7x8+73x8+...+799x8
M = 8x(7+73+...+799)
mà 8 chia hết 8 => 8(7+73+...+799) chia hết 8
Vậy M chia hết cho 8
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Mấy bài thực hiện phép tính bn tự làm nha
3.|x+4|-2.(x-1)=7-2x
3.|x-4|-2x-2=7-2x
=>2.|x-4| =7-2x+2x+2
=>2.|x-4| =9-4x
=>|x-4| =4,5-2x
=>x-4=4,5-2x hoặc x-4=-4,5+2x
2+2x=4,5+4 4,5-4=2x-2
3x =8,5 0,5 =x
x =8,5:3
x =2,833
Vậy x\(\in\){2,833;0,5}
Phần còn lại bn làm tương tự nha
Chúc bn học tốt
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\(\Delta'=9-6m+m^2=\left(m-3\right)^2\ge0;\forall m\)
\(\Rightarrow\) Pt luôn có nghiệm với mọi m
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-6\\x_1x_2=6m-m^2\end{matrix}\right.\)
Do \(x_1\) là nghiệm nên: \(x_1^2+6x_1+6m-m^2=0\Leftrightarrow2x_1^2+12x_1=2m^2-12\)
\(x_1^3-x_2^3+2x_1^2+12x_1+72=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left[\left(x_1+x_2\right)^2-x_1x_2\right]+2m^2-12m+72=0\)
\(\Leftrightarrow\left(x_1-x_2\right)\left(m^2-6m+36\right)+2m^2-12m+72=0\)
\(\Leftrightarrow\left(x_1-x_2+2\right)\left(m^2-6m+36\right)=0\)
\(\Leftrightarrow x_1-x_2+2=0\) (do \(m^2-6m+36=\left(m-3\right)^2+27>0;\forall m\))
Kết hợp với \(x_1+x_2=-6\) ta được:
\(\left\{{}\begin{matrix}x_1-x_2=-2\\x_1+x_2=-6\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=-4\\x_2=-2\end{matrix}\right.\)
Thế vào \(x_1x_2=6m-m^2\)
\(\Rightarrow8=6m-m^2\Rightarrow m^2-6m+8=0\Rightarrow\left[{}\begin{matrix}m=2\\m=4\end{matrix}\right.\)
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a) \(x^2+2\left(m-1\right)x-6m-7=0\)\(0\)
\(\left(a=1;b=2\left(m-1\right);b'=m-1;c=-6m-7\right)\)
\(\Delta'=b'^2-ac\)
\(=\left(m-1\right)^2-1.\left(-6m-7\right)\)
\(=m^2-2m+1+6m+7\)
\(=m^2+4m+8\)
\(=m^2+2.m.2+2^2+4\)
\(=\left(m+2\right)^2+4>0,\forall m\)
Vì \(\Delta'>0\) nên phương trình ( 1 ) luôn có 1 nghiệm phân biệt với mọi m
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\(1,\Leftrightarrow7x=42\Leftrightarrow x=6\\ 2,\Leftrightarrow36-4x=4\Leftrightarrow4x=32\Leftrightarrow x=8\\ 3,\Leftrightarrow3x=35\Leftrightarrow x=\dfrac{35}{3}\\ 4,\Leftrightarrow x-12=144\Leftrightarrow x=156\\ 5,\Leftrightarrow x-14=16\Leftrightarrow x=30\\ 6,\Leftrightarrow3x-24=\dfrac{148}{73}\Leftrightarrow3x=\dfrac{1900}{73}\Leftrightarrow x=\dfrac{1900}{219}\\ 7,\Leftrightarrow33+x=45\Leftrightarrow x=12\\ 8,Sai.đề\\ 9,\Leftrightarrow\left(x+9\right):2=39\Leftrightarrow x+9=78\Leftrightarrow x=69\\ 11,\Leftrightarrow2\left(x+7\right)=38\Leftrightarrow x+7=19\Leftrightarrow x=12\\ 13,\Leftrightarrow2\left(x-51\right)=66\Leftrightarrow x-51=33\Leftrightarrow x=84\)
\(15,\Leftrightarrow x-19=9\Leftrightarrow x=28\\ 17,\Leftrightarrow\left(x-3\right):2=48\Leftrightarrow x-3=96\Leftrightarrow x=99\\ 19,\Leftrightarrow0x=46\Leftrightarrow x\in\varnothing\\ 8,Sai.đề\\ 14,\Leftrightarrow2x=209\Leftrightarrow x=\dfrac{209}{2}\\ 16,\Leftrightarrow2x+6=157\Leftrightarrow2x=151\Leftrightarrow x=\dfrac{151}{2}\\ 18,\Leftrightarrow5\left(x+4\right)=100\Leftrightarrow x+4=20\Leftrightarrow x=16\\ 20,\Leftrightarrow\left(3x-5\right)^3=27=3^3\Leftrightarrow3x-5=3\Leftrightarrow x=\dfrac{8}{3}\)
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a: =>7x=42
hay x=6
b: =>5x=35
hay x=7
c: =>x-14=16
hay x=30
d: =>36-4x=4
=>4x=32
hay x=8
e: =>x-12=144
hay x=156
f: =>3x-16=14
hay x=10
g: =>x+33=45
hay x=12
h: =>(x+9):2=39
=>x+9=78
hay x=69
a: =>7x=42
hay x=6
b: =>5x=35
hay x=7
c: =>x-14=16
hay x=30
d: =>36-4x=4
=>4x=32
\(M=7+7^2+7^3+...+7^{2021}\)
\(7M=7^2+7^3+7^4+...+7^{2022}\)
\(7M-M=\left(7^2+7^3+7^4+...+7^{2022}\right)-\left(7+7^2+7^3+...+7^{2021}\right)\)
\(6M=7^2+7^3+7^4+...+7^{2022}-7-7^2-7^3-...-7^{2021}\)
\(6M=7^{2022}-7\)
Lại có :
\(6M+7=7x\)
\(\Leftrightarrow7^{2022}-7+7=7x\)
\(7^{2022}=7x\)
\(7x=7^{2022}\)
\(x=7^{2022}:7\)
\(x=7^{2021}\)