) B = (x+2) (x+3)(x+4)(x+5)-24
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Bài làm:
a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)
\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)
Đặt \(x^2+5x+5=t\)\(\Rightarrow\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2\)
\(=\left(x^2+5x+5\right)^2\)
b) Tương tự như a phân tích và đặt ra được: \(t^2-1-24=t^2-25=\left(t-5\right)\left(t+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)=x\left(x+5\right)\left(x^2+5x+10\right)\)
c) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)
\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
Đặt \(x^2+8x+11=t\)\(\Rightarrow\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1\)
\(=\left(t-1\right)\left(t+1\right)=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)
\(=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)
d) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+11=t\)\(\Rightarrow\left(t-1\right)\left(t+1\right)-24=t^2-1-24=t^2-25\)
\(=\left(t-5\right)\left(t+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
Làm mẫu cho 1 vd:
a, (x+1)(x+2)(x+3)(x+4)+1
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)(1)
Đặt \(y=x^2+5x+5\)
Khi đó ::
(1) = \(\left(y-1\right)\left(y+1\right)+1\)
\(=y^2-1+1=y^2\)
Thay vào ta được: \(\left(x^2+5x+5\right)^2\)
\(a)x=\dfrac{1}{4}+\dfrac{5}{13}=\dfrac{33}{52}.\\ b)\dfrac{x}{3}=\dfrac{2}{3}+\dfrac{-1}{7}.\\ \Leftrightarrow\dfrac{x}{3}=\dfrac{11}{21}.\\ \Leftrightarrow\dfrac{7x}{21}=\dfrac{11}{21}.\\ \Rightarrow7x=11.\\ \Leftrightarrow x=\dfrac{11}{7}.\\ c)\dfrac{x}{3}=\dfrac{16}{24}+\dfrac{24}{36}=\dfrac{2}{3}+\dfrac{2}{3}=\dfrac{4}{3}.\\ \Rightarrow x=4.\\ d)\dfrac{x}{15}=\dfrac{1}{5}+\dfrac{2}{3}=\dfrac{13}{15}.\\ \Rightarrow x=13.\)
a: =>(x-2)^3*[(x-2)^8-1]=0
=>(x-2)(x-3)(x-1)=0
=>\(x\in\left\{2;3;1\right\}\)
b: (x-5)^24=(x-5)^9
=>\(\left(x-5\right)^9\cdot\left[\left(x-5\right)^{15}-1\right]=0\)
=>x-5=0 hoặc x-5=1
=>x=6 hoặc x=5
c: =>(x-5)^4*[(x-5)^21-1]=0
=>x-5=0 hoặc x-5=1
=>x=5 hoặc x=6
a) \(\left(x-2\right)^{11}=\left(x-2\right)^3\)
\(\Rightarrow\left(x-2\right)^{11}-\left(x-2\right)^3=0\)
\(\Rightarrow\left(x-2\right)^3\left[\left(x-2\right)^8-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^3=0\\\left(x-2\right)^8-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\\left(x-2\right)^8=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x-2=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
b) \(\left(x-5\right)^{24}=\left(x-5\right)^9\)
\(\Rightarrow\left(x-5\right)^{24}-\left(x-5\right)^9=0\)
\(\Rightarrow\left(x-5\right)^9\left[\left(x-5\right)^{15}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^9=0\\\left(x-5\right)^{15}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{15}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
c) \(\left(x-5\right)^{25}=\left(x-5\right)^4\)
\(\Rightarrow\left(x-5\right)^{25}-\left(x-5\right)^4\)
\(\Rightarrow\left(x-5\right)^4\left[\left(x-5\right)^{21}-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-5\right)^4=0\\\left(x-5\right)^{21}-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-5=0\\\left(x-5\right)^{21}=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x-5=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)
a, \(390-\left(x-7\right)=13^2:12\)
\(390-\left(x-7\right)=\) \(\dfrac{169}{12}\)
\(x-7=390-\dfrac{169}{12}\)
\(x-7=\dfrac{4511}{12}\)
\(x=\dfrac{4511}{12}+7\)
\(x=\dfrac{4595}{12}\)
Vậy ...
b, \(\left(x-35.2^2\right):7=3^3-24\)
\(\left(x-35.4\right):7=27-24\)
\(\left(x-140\right):7=3\)
\(\Leftrightarrow\left(x-140\right)=3.7\)
\(\Leftrightarrow x-140=21\)
\(\Leftrightarrow x=161\)
Vậy .....
c) \(x-6:2-\left(4^2.3-24\right):2:6=3\)
\(x-3-\left(16.3-24\right):2:6=3\)
\(x-3-\left(48-24\right):2:6=3\)
\(x-3-24:2:6=3\)
\(x-3-2=3\)
\(x=3+2+3\)
\(x=8\)
Vậy ......
d) \(4x-5=5+5^2+5^3+.....+5^{99}\)
Đặt :
\(A=5+5^2+.........+5^{99}\)
\(\Leftrightarrow5A=5^2+5^3+..........+5^{100}\)
\(\Leftrightarrow5A-A=\left(5^2+5^3+......+5^{100}\right)-\left(5+5^2+....+5^{99}\right)\)
\(\Leftrightarrow4A=5^{100}-5\)
\(\Leftrightarrow A=\dfrac{5^{100}-5}{4}\)
\(\Leftrightarrow4x+5=\dfrac{5^{100}-5}{4}\)
Đến đây thì sao nữa nhỉ ?
e) \(\left(2x-1\right)^4=625\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(2x-1\right)^4=5\\\left(2x-1\right)^4=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=5\\2x-1=-5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
Vậy ....
\(\dfrac{3}{7}\times\dfrac{7}{9}\times\dfrac{1}{2}\)
\(=\dfrac{3\times7\times1}{7\times9\times2}\)
\(=\dfrac{21}{126}\)
\(=\dfrac{1}{6}\)
\(\dfrac{5}{8}\times4\times\dfrac{1}{2}\\ =\dfrac{5}{8}\times\dfrac{4}{1}\times\dfrac{1}{2}\\ =\dfrac{5\times4\times1}{8\times1\times2}\\ =\dfrac{20}{16}\\ =\dfrac{5}{4}\)
\(4\times\dfrac{1}{24}\times3\\ =\dfrac{4}{1}\times\dfrac{1}{24}\times\dfrac{3}{1}\\ =\dfrac{4\times1\times3}{1\times24\times1}\\ =\dfrac{12}{24}\\ =\dfrac{1}{2}\)
nhiều quá :((
\(a,2\left(x-5\right)-3\left(x+7\right)=14\)
\(2x-10-3x-21=14\)
\(-x-31=14\)
\(-x=45\)
\(x=45\)
\(b,5\left(x-6\right)-2\left(x+3\right)=12\)
\(5x-30-2x-6=12\)
\(3x-36==12\)
\(3x=48\)
\(x=16\)
\(c,3\left(x-4\right)-\left(8-x\right)=12\)
\(3x-12-8+x=0\)
\(4x-20=0\)
\(4x=20\)
\(x=5\)
Cố nốt nha bn !
cảm ơn, bn nha:)))
mà hình như bạn TOP 3 trả lời câu hỏi pải ko nhỉ???
a) \(\frac{24}{x}:\frac{8}{3}=\frac{3}{5}\) \(\frac{24}{x}=\frac{3}{5}.\frac{8}{3}\) \(\frac{24}{x}=\frac{8}{5}\) \(x=24.5:8\) \(x=15\)
b) Đề bài sai rồi
\(B=\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)