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\(\left(x+1\right)^{x+2}=\left(x+1\right)^{x+4}\)
\(\Rightarrow x+2=x+4\)
\(\Rightarrow0x=2\)
=> không có giá trị của x thỏa mãn
=.= hk tốt!!
(x + 1)x + 2 = (x + 1)x + 4
<=> x + 2 = x + 4
<=> 2 = x + 4 - x
<=> 2 = 4
<=> 0 = 4 - 2
<=> 0 = 2
=> không có x thỏa mãn đề bài
13+23+33+...+1003
=1+2+1.2.3+3+2.3.4+100+99.100.101
=(1+2+3+...+100)+(1.2.3+2.3.4+...+99.100.101)
=5050+101989800
=101994850
NHỚ T.I.C.K và KB với mk nha
Ta có:
\(\left|\frac{-1}{2}\right|:x=\frac{-1}{2}\)
\(\Rightarrow\frac{1}{2}:x=\frac{-1}{2}\)
\(\Rightarrow x=\frac{1}{2}:\left(\frac{-1}{2}\right)\)
\(\Rightarrow x=-1\)
Vậy \(x=-1\)
#Mạt Mạt#
\(\left|-\frac{1}{2}\right|^3:x=-\frac{1}{2}\)
\(\left(\frac{1}{2}\right)^3:x=-\frac{1}{2}\)
\(\frac{1}{8}:x=-\frac{1}{2}\)
\(x=\frac{1}{8}:\left(-\frac{1}{2}\right)\)
\(x=-\frac{1}{4}\)
x2 + 4x + 3
<=> 2x2 - 3x - x + 3
<=> (x2 - 3x) - (x - 3)
<=> x.(x - 3) - (x - 3)
<=> (x - 1)(x - 3) = 0
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=3\end{cases}}\)
Vậy:..
a) \(5^{-1}.25^x=125\)
\(\Rightarrow5^{-1}.5^{2x}=5^3\)
\(\Rightarrow5^{2x-1}=5^3\)
\(\Rightarrow2x-1=3\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
b) \(|x+1|+|x+2|+|x+3|=4x\)
Vì \(\hept{\begin{cases}|x+1|\ge0\forall x\\|x+2|\ge0\forall x\\|x+3|\ge0\forall x\end{cases}}\)
\(\Rightarrow|x+1|+|x+2|+|x+3|\ge0\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow x\ge0\)
\(\Rightarrow\hept{\begin{cases}x+1>0\\x+2>0\\x+3>0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}|x+1|=x+1\\|x+2|=x+2\\|x+3|=x+3\end{cases}}\)
\(\Rightarrow\left(x+1\right)+\left(x+2\right)+\left(x+3\right)=4x\)
\(\Rightarrow3x+6=4x\)
\(\Rightarrow x=6\)
Vậy \(x=6\)
Câu 1 : Đặt A = 1.2.3 + 2.3.4 + ... + 111.112.113
=> 4A = 1.2.3.4 + 2.3.4.4 + ... + 111.112.113.4
= 1.2.3.4 + 2.3.4.(5 - 1) + .... + 111.112.113.(114 - 110)
= 1.2.34 + 2.3.4.5 - 1.2.3.4 + ... + 111.112.113.114 - 110.111.112.113
= 111.112.113.114
=> A = 111.113.114.28 = 40 037 256
Câu 2 Đặt A = 1.2 + 2.3 + 3.4 + ... + 277.278
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 277.278.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 277.278.(279 - 276)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 277.278.279 - 276.277.278
= 277.278.279
=> A = 7161558
3) Đặt A = 1.4 + 2.5 + ... + 277.280
= 1.(2 + 2) + 2.(2 + 3) + ... + 277.(278 + 2)
= (1.2 + 2.3 + .... + 277.278) + 2(1 + 2 + .... 277)
Đặt B = 1.2 + 2.3 + .... + 277.278
=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + ... + 277.278.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 277.278.(279 - 276)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 277.278.279 - 276.277.278
= 277.278.279
=> B = 7161558
Khi đó A = B + 2(1 + 2 + .... 277)
= 7161558 + 2.277(277 + 1) : 2
= 7238564
Câu 4 : \(\left(\frac{2^2}{2.4}+\frac{2^2}{4.6}+...+\frac{2^2}{34.36}\right)x-1\frac{1}{6}=1\frac{2}{3}\)
=> \(2\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{34.36}\right)x-\frac{7}{6}=\frac{5}{3}\)
=> \(2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{34}-\frac{1}{36}\right)x=\frac{17}{6}\)
=> \(\left(\frac{1}{2}-\frac{1}{36}\right)x=\frac{17}{12}\)
=> x = 3
Câu 5 : Đặt A = 1 + 2 + 22 + ... + 29 (1)
=> 2A = 2 + 22 + 23 + ... + 210 (2)
Lấy (2) trừ (1) theo vế ta có :
2A - A = (2 + 22 + 23 + ... + 210) - ( 1 + 2 + 22 + ... + 29)
A = 210 - 1 = 1024 - 1 = 1023
Câu 6 : Đặt A = 12 + 22 + 32 + .... + 1002
= 1.1 + 2.2 + 3.3 + ... + 100.100
= 1.(2 - 1) + 2(3 - 1) + 3(4 - 1) + ... + 100(101 - 1)
= (1.2 + 2.3 + 3.4 + ... + 100.101) - (1 + 2 + 3 + 4 + ... + 100)
Đặt B = 1.2 + 2.3 + 3.4 + ... + 100.101
=> 3B = 1.2.3 + 2.3.3 + 3.4.3 + ... + 100.101.3
= 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 100.101(102 - 99)
= 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + .... + 100.101.102 - 99.100.101
= 100.101.102
=> B = 343400
Khi đó A = B - (1 + 2 + 3 + 4 + ... + 100)
= 343 400 - [100.(100 + 1) : 2]
= 338 350
\(\left(x-1\right)^2=\left(x-1\right)^4\)
\(\Rightarrow\left(x-1\right)^2-\left(x-1\right)^4=0\)
\(\Rightarrow\left(x-1\right)^2.1-\left(x-1\right)^2.\left(x-1\right)^2=0\)
\(\Rightarrow\left(x-1\right)^2.\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=0\\\left(x-1\right)^2-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\\left(x-1-1\right)\left(x-1+1\right)=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=2;x=0\end{cases}}}\)
Vậy: \(x\in\left\{1;2;0\right\}\)
(x - 1)2 = (x - 1)4
<=> (x - 1)2 - (x - 1)4 = 0
<=> (x - 1)2 - (x - 1)2.(x - 1)2 = 0
<=> (x - 1)2. [1 - (x - 1)2] = 0
<=> x - 1 = 0 hoặc 1 - (x - 1)2 = 0
<=> x = 1 <=> (x - 1)2 = 1
<=> x - 1 = 1 hoặc x - 1 = -1
<=> x = 2 <=> x = 0
Vậy x = 1; x = 2; x = 0