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a) \(125\cdot\left(-24\right)+24\cdot225\)
\(=\left(225-125\right)\cdot24\)
\(=100\cdot24\)
\(=2400\)
b) \(26\cdot\left(-125\right)-125\cdot\left(-36\right)\)
\(=\left(36-26\right)\cdot125\)
\(=10\cdot125\)
\(=1250\)
a) \(\left(2.x+1\right)^3=125\)
\(\Leftrightarrow\left(2.x+1\right)^3=5^3\)
\(\Leftrightarrow2.x+1=5\)
\(2.x=5-1\)
\(2.x=4\)
\(x=4:2\)
\(x=2\)
b) \(\left(4.x-1\right)^2=25.9\)
\(\left(4.x-1\right)^2=225\)
\(\Leftrightarrow\left(2.x-1\right)^2=15^2\)
\(\Leftrightarrow2.x-1=15\)
\(\Rightarrow2.x=15+1\)
\(\Rightarrow2.x=16\)
\(\Rightarrow x=16:2\)
\(\Rightarrow x=8\)
a,(2x+1)3=53
2x+1=5
2x=4
x=2
b,(4x-1)2=225
(4x-1)2=152
4x-1=15
4x=16
x=4
Bài 2:
a) \(\left(x-3\right)^3+27=0\)
\(\Leftrightarrow\left(x-3\right)^3=0-27\)
\(\Leftrightarrow\left(x-3\right)^3=-27\)
\(\Leftrightarrow\left(x-3\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x-3=-3\)
\(\Leftrightarrow x=\left(-3\right)+3\)
\(\Leftrightarrow x=0\)
b) \(-125-\left(x+1\right)^3=0\)
\(\Leftrightarrow\left(x+1\right)^3=-125-0\)
\(\Leftrightarrow\left(x+1\right)^3=-125\)
\(\Leftrightarrow\left(x+1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow x+1=-5\)
\(\Leftrightarrow x=\left(-5\right)-1\)
\(\Leftrightarrow x=-6\)
c) \(\left(2x-\dfrac{1}{4}\right)^2-\dfrac{1}{16}=0\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=0+\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Leftrightarrow2x-\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{4}+\dfrac{1}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{2}:2\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
d) \(2^x+2^{x+1}=24\)
\(\Leftrightarrow2^x+2^x.2=24\)
\(\Leftrightarrow2^x\left(1+2\right)=24\)
\(\Leftrightarrow2^x.3=24\)
\(\Leftrightarrow2^x=24:3\)
\(\Leftrightarrow2^x=8\)
\(\Leftrightarrow2^x=2^3\)
\(\Rightarrow x=3\)
e) \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=1\)
\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=1+\dfrac{1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=-\dfrac{3}{2}\\x+\dfrac{1}{5}=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)
g) \(\left|x-3\right|+2x=10\)
\(\Leftrightarrow\left|x-3\right|=10-2x\)
\(\Leftrightarrow\left|x-3\right|=2.5-2x\)
\(\Leftrightarrow\left|x-3\right|=2\left(5-x\right)\)
(không chắc có nên làm tiếp câu g không, thấy đề cứ là lạ, có j sai sai...)
Bài 1:
a) \(2^7+2^9⋮10\)
Ta có: \(2^7+2^9=2^{4.1}.2^3+2^{4.2}.2\)
\(\Leftrightarrow\overline{A6}.2^3+\overline{B6}.2\)
\(\Leftrightarrow\overline{A6}.8+\overline{B6}.2\)
\(\Leftrightarrow\overline{C8}+\overline{D2}\)
\(\Leftrightarrow\overline{E0}\)
Mà \(\overline{E0}⋮10\) \(\Rightarrow2^7+2^9⋮10\)
b) \(8^{24}.25^{10}⋮2^{36}.5^{20}\)
Ta có: \(8^{24}.25^{10}=\left(2^3\right)^{24}.\left(5^2\right)^{10}\)
\(\Leftrightarrow2^{72}.5^{20}\)
Do \(2^{72}⋮2^{36}\) và \(5^{20}⋮5^{20}\) \(\Rightarrow8^{24}.25^{10}⋮2^{36}.5^{20}\)
c) \(3^{10}+3^{12}⋮30\)
Ta có: \(3^{10}+3^{12}=3^{4.2}.3^2+3^{4.3}\)
\(\Leftrightarrow\overline{A1}.3^2+\overline{B1}\)
\(\Leftrightarrow\overline{A1}.9+\overline{B1}\)
\(\Leftrightarrow\overline{C9}+\overline{B1}\)
\(\Leftrightarrow\overline{D0}⋮10\)
(Chứng minh chia hết cho 10 rồi chứng minh chia hết cho 3, mình chưa tìm được cách làm, chờ chút)
a ) ( x + 1 ) x ( x2 - 4 ) = 0
vậy chắc chắn 1 biểu thức phải bằng 0 để có kết quả đúng . vậy chỉ có thể là x2 - 4 = 0
vì phép còn lại là x + 1 = số nguyên dương
x2 - 4 = 0
x = 2
b ) x15 = x
vậy quá rõ x = 1 , 0
vì chỉ có 2 số này nhân bao nhiêu lần chính nó cũng bằng nó
c ) ( x - 5 ) 4 = ( x - 5 )6
4 x - 625 = 6 x - 15625
4 x + 15625 - 625 = 6 x
4 x + 15000 = 6 x
15000 = 2 x
x = 7500
d ) làm sau
a. \(\left(x+1\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)\left(x+2\right)=0\)
TH1: \(x+1=0\Rightarrow x=-1\)
TH2: \(x-2=0\Rightarrow x=2\)
TH3: \(x+2=0\Rightarrow x=-2\)
Vậy:...
b) \(x^{15}=x\)
\(\Rightarrow x\in\left\{0;1;-1\right\}\)
c) \(\left(x-5\right)^4=\left(x-5\right)^6\)
TH1:\(x-5=1\Rightarrow x=6\)
TH2: \(x-5=-1\Rightarrow x=4\)
TH3: \(x-5=0\Rightarrow x=5\)
d) \(\left(2x+1\right)^3=125\)
\(\Leftrightarrow2x+1=\sqrt[3]{125}=5\)
\(\Leftrightarrow x=2\)
a x+35=515/5=103
x=103-35=68
b 3(x+1)=96-42=54
x+1=54/3=18
x=18-1=7
a) \(5\left(x+35\right)=515\)
\(\Rightarrow x+35=103\)
\(\Rightarrow x=68\)
b) \(96-3\left(x+1\right)=42\)
\(\Rightarrow3\left(x+1\right)=54\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=17\)
c) \(5^x.5=5^4\Rightarrow5^x=5^3\Rightarrow x=3\)
d) \(\left(x-1\right)^2=125\)
Mà \(\orbr{\begin{cases}\left(5\sqrt{5}\right)^2=125\\\left(-5\sqrt{5}\right)^2=125\end{cases}\Rightarrow\orbr{\begin{cases}x-1=5\sqrt{5}\\x-1=-5\sqrt{5}\end{cases}\Rightarrow}\orbr{\begin{cases}x=5\sqrt{5}+1\\x=1-5\sqrt{5}\end{cases}}}\)
Mà lớp 6 chưa học căn
=> Kiểm tra lại đề
Bài 1:
\(a.\left(-356+57\right)-\left(27-356\right)=-356+57-27+356=\left(-356+356\right)+\left(57-27\right)=30\) \(b.125.\left(-24+24.225\right)=125.\left(-24+5400\right)=125.\left(-24\right)+125.5400=-3000+675000=672000\)
\(c.26.\left(-125\right)-125.\left(-36\right)=-125.\left(26-36\right)=-125.\left(-10\right)=1250\)
Bài 2:
\(a.\left(2x-4\right)^2=0\)
\(\Rightarrow2x-4=0\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
\(b.\frac{x+5}{x+3}=\frac{x+3+2}{x+3}=\frac{x+3}{x+3}+\frac{2}{x+3}=1+\frac{2}{x+3}\)
Để (x+5) chia hết cho (x+3) thì 2 phải chia hết cho (x+3)
\(\Rightarrow x+3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(x+3=1\Rightarrow x=-2\)
\(x+3=-1\Rightarrow x=-4\)
\(x+3=2\Rightarrow x=-1\)
\(x+3=-2\Rightarrow x=-5\)
Vậy \(x\in\left\{-2;-4;-1;-5\right\}\)
Bài 2:
a)\(\left(2x-4\right)^2=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\Leftrightarrow x=2\)
b)\(\frac{x+5}{x+3}=\frac{x+3+2}{x+3}=\frac{x+3}{x+3}+\frac{2}{x+3}=1+\frac{2}{x+3}\in Z\)
Suy ra \(2⋮x+3\Rightarrow x+3\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
\(\Rightarrow x\in\left\{-2;-4;-1;-5\right\}\)