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a: \(2^7+\left(x-3^7\right)=5^7-4^7\)
=>\(128+x-2187=78125-16384\)
=>\(x-2059=61741\)
=>\(x=61741+2059=63800\)
c: \(7^2-\left(x+15\right)=5\cdot2^2\)
=>49-(x+15)=5*4=20
=>x+15=29
=>x=14
d: 7^3-7(13-x)=14
=>343-7(13-x)=14
=>7(13-x)=343-14=329
=>13-x=47
=>x=13-47=-34
cậu giải thích giùm mình đoạn này với P(x)=x^7-(x+1)x^6+(x+1)x^5-(x+1)x^4+(x+1)x^3-(x+1)x^2+(x+1)x+15
P(x)=x^7-x^7-x^6+x^6+x^5-x^5-x^4+x^4+x^3-x^3-x^2+x^2+x+15
P(x)=x+15=79+15=94
hay giai giup mk may phan nay nhe
cmr cac bieu thuc sau ko phu thuoc vao x:
c)C=x(x^3+x^2-3x-2)-(x^2-2)(x^2+x-1)
e)E=(x+1)(x^2-x+1)-(x-1)(x^2+x+1)
tinh gia tri cua da thuc
b)Q(x)=x^14-10x^13=10x^12-10x^11+...+10x^2-10x+10 voi x=9
c)R(x)=x^4-17x^3+17x^2_17x+20 või=16
d)S(x)=x^10-13x^9+13x^8-13X^7+...+13x^2-13x+10 voi 12
Bài 1:
a) \(=\dfrac{8}{15}\left(\dfrac{7}{13}+\dfrac{6}{13}\right)=\dfrac{8}{15}.1=\dfrac{8}{15}\)
b) \(=\dfrac{3.3-7-2.4}{12}=-\dfrac{6}{12}=-\dfrac{1}{2}\)
Bài 2:
\(\dfrac{x}{2,7}=-\dfrac{2}{3,6}\Rightarrow x=\dfrac{\left(-2\right).2,7}{3,6}\Rightarrow x=-\dfrac{3}{2}\)
Bài 3:
\(\dfrac{x}{2}=\dfrac{y}{5}=\dfrac{x+y}{2+5}=-\dfrac{21}{7}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).2=-6\\y=\left(-3\right).5=-10\end{matrix}\right.\)
`@` `\text {Ans}`
`\downarrow`
`B(x)-A(x)+C(x)`
`=`\((x^2-5x^3-4x+7) - (-x^3 + 7x^2 +2x - 15) + 3x^3 - 7x^2 -4\)
`=`\(x^2-5x^3-4x+7+x^3-7x^2-2x+15+3x^3-7x^2-4\)
`=`\(\left(-5x^3+x^3+3x^3\right)+\left(x^2-7x^2-7x^2\right)+\left(-4x-2x\right)+\left(7+15-4\right)\)
`=`\(-x^3-13x^2-6x+18\)
`C(x)-B(x)-A(x)`
`=`\(3x^3 - 7x^2 -4 - (x^2-5x^3-4x+7) - (-x^3 + 7x^2 +2x - 15)\)
`=`\(3x^3-7x^2-4-x^2+5x^3+4x-7+x^3-7x^2-2x+15\)
`=`\(\left(3x^3+5x^3+x^3\right)+\left(-7x^2-x^2-7x^2\right)+\left(4x-2x\right)+\left(-4-7+15\right)\)
`=`\(9x^3-15x^2+2x+4\)
a) \(B\left(x\right)-A\left(x\right)+C\left(x\right)\)
\(=\left(x^2-5x^3-4x+7\right)-\left(-x^3+7x^2+2x-15\right)+\left(3x^3-7x^2-4\right)\)
\(=x^2-5x^3-4x+7+x^3-7x^2-2x+15+3x^3-7x^2-4\)
\(=\left(-5x^3+x^3+3x^3\right)+\left(x^2-7x^2-7x^2\right)-\left(4x+2x\right)+\left(7-4+15\right)\)
\(=-x^3-13x^2-6x+18\)
b) \(C\left(x\right)-B\left(x\right)-A\left(x\right)\)
\(=\left(3x^3-7x^2-4\right)-\left(x^2-5x^3-4x+7\right)-\left(-x^3+7x^2+2x-15\right)\)
\(=3x^3-7x^2-4-x^2+5x^3+4x-7+x^3-7x^2-2x+15\)
\(=\left(3x^3+5x^3+x^3\right)-\left(7x^2+x^2+7x^2\right)+\left(4x-2x\right)-\left(4+7-15\right)\)
\(=9x^3-15x^2+2x+4\)
a, (\(\dfrac{9}{10}\) - \(\dfrac{15}{16}\)) \(\times\) ( \(\dfrac{5}{12}\) - \(\dfrac{11}{15}\) - \(\dfrac{7}{20}\))
= (\(\dfrac{72}{80}\) - \(\dfrac{75}{80}\)) \(\times\) (\(\)\(\dfrac{25}{60}\) - \(\dfrac{44}{60}\) - \(\dfrac{21}{60}\))
= - \(\dfrac{3}{80}\) \(\times\) (- \(\dfrac{2}{3}\))
= \(\dfrac{1}{40}\)
b, (-1)3 + (- \(\dfrac{2}{3}\))2 : 2\(\dfrac{2}{3}\) + \(\dfrac{5}{6}\)
= -13 + \(\dfrac{4}{9}\) : \(\dfrac{8}{3}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{4}{9}\) \(\times\) \(\dfrac{3}{8}\) + \(\dfrac{5}{6}\)
= -1 + \(\dfrac{1}{6}\) + \(\dfrac{5}{6}\)
= -1 + 1
= 0
a) \(\frac{x}{6}=\frac{8}{3}\Rightarrow x=\frac{8\cdot6}{3}=16\)
b) \(\frac{-3}{x}=\frac{15}{7}\Rightarrow x=-\frac{3\cdot7}{15}=-\frac{7}{5}\)
c) \(\frac{0,1}{5}=\frac{x}{15}\Rightarrow x=\frac{15\cdot0,1}{5}=0,3\)
d) \(\frac{x-3}{3}=\frac{4}{5}\Leftrightarrow x-3=\frac{4\cdot3}{5}=\frac{12}{5}\)
\(\Rightarrow x=\frac{12}{5}+3=\frac{27}{5}\)
a) \(\dfrac{2}{3}\left(x+1\right)-\dfrac{4}{5}\left(x+2\right)=\dfrac{35}{2}\)
\(\Rightarrow\dfrac{2}{3}x+\dfrac{2}{3}-\dfrac{4}{5}x-\dfrac{8}{5}=\dfrac{35}{2}\)
\(\Rightarrow\left(\dfrac{2}{3}-\dfrac{4}{5}\right)x+\left(\dfrac{2}{3}-\dfrac{8}{5}\right)=\dfrac{35}{2}\)
\(\Rightarrow-\dfrac{2}{15}x-\dfrac{14}{15}=\dfrac{35}{2}\)
\(\Rightarrow-\dfrac{2}{15}x=\dfrac{553}{30}\)
\(\Rightarrow x=\dfrac{553}{30}:-\dfrac{2}{15}\)
\(\Rightarrow x=-\dfrac{553}{4}\)
b) \(4\left(x-2\right)+5\left(x+1\right)=-15\)
\(\Rightarrow4x-8+5x+5=-15\)
\(\Rightarrow\left(4+5\right)x+\left(-8+5\right)=-15\)
\(\Rightarrow9x-3=-15\)
\(\Rightarrow9x=-15+3\)
\(\Rightarrow x=\dfrac{-12}{9}\)
\(\Rightarrow x=-\dfrac{4}{3}\)
c) \(\dfrac{3}{2}:x+\left(-\dfrac{5}{2}\right)=-\dfrac{7}{3}\)
\(\Rightarrow\dfrac{3}{2}:x=-\dfrac{7}{3}+\dfrac{5}{2}\)
\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{6}\)
\(\Rightarrow x=\dfrac{1}{6}:\dfrac{3}{2}\)
\(\Rightarrow x=\dfrac{1}{9}\)
Bài 4:
a) \(\dfrac{4}{3}+\left(1,25-x\right)=2,25\)
\(1,25-x=2,25-\dfrac{4}{3}=\dfrac{9}{4}-\dfrac{4}{3}\)
\(1,25-x=\dfrac{11}{12}\)
\(x=1,25-\dfrac{11}{12}=\dfrac{5}{4}-\dfrac{11}{12}\)
\(x=\dfrac{1}{3}\)
b) \(\dfrac{17}{6}-\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(x-\dfrac{7}{6}=\dfrac{17}{6}-\dfrac{7}{4}=\dfrac{34}{12}-\dfrac{21}{12}\)
\(x-\dfrac{7}{6}=\dfrac{13}{12}\)
\(x=\dfrac{13}{12}+\dfrac{7}{6}=\dfrac{13}{12}+\dfrac{14}{12}\)
\(x=\dfrac{27}{12}=\dfrac{9}{4}\)
c) \(4-\left(2x+1\right)=3-\dfrac{1}{3}=\dfrac{9}{3}-\dfrac{1}{3}\)
\(4-\left(2x+1\right)=\dfrac{8}{3}\)
\(2x+1=\dfrac{8}{3}+4=\dfrac{8}{3}+\dfrac{12}{3}\)
\(2x+1=\dfrac{20}{3}\)
\(2x=\dfrac{20}{3}-1=\dfrac{20}{3}-\dfrac{3}{3}\)
\(2x=\dfrac{17}{3}\)
\(x=\dfrac{17}{3}.\dfrac{1}{2}=\dfrac{17}{6}\)
Bài 15:
a) \(\left(\dfrac{-2}{3}\right)^9:x=\dfrac{-2}{3}\)
\(x=\left(\dfrac{-2}{3}\right)^9:\dfrac{-2}{3}=\left(\dfrac{-2}{3}\right)^{9-1}\)
\(=>x=\left(\dfrac{-2}{3}\right)^8\)
b) \(x:\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^4\)
\(x=\left(\dfrac{4}{9}\right)^4.\left(\dfrac{4}{9}\right)^5=\left(\dfrac{4}{9}\right)^{4+5}\)
\(=>x=\left(\dfrac{4}{9}\right)^9\)
c) \(\left(x+4\right)^3=-125\)
\(\left(x+4\right)^3=\left(-5\right)^3\)
\(=>x+4=-5\)
\(x=-5-4\)
\(=>x=-9\)
d) \(\left(10-5x\right)^3=64\)
\(\left(10-5x\right)^3=4^3\)
\(=>10-5x=4\)
\(5x=10-4\)
\(5x=6\)
\(=>x=\dfrac{6}{5}\)
e) \(\left(4x+5\right)^2=81\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(4x+5\right)^2=\left(-9\right)^2\\\left(4x+5\right)^2=9^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x+5=-9\\4x+5=9\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=-14\\4x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-14}{4}\\x=1\end{matrix}\right.\)
Bài 16:
a) \(4-1\dfrac{2}{5}-\dfrac{8}{3}\)
\(=4-\dfrac{7}{5}-\dfrac{8}{3}\)
\(=\dfrac{60-21-40}{15}=\dfrac{-1}{15}\)
b) \(-0,6-\dfrac{-4}{9}-\dfrac{16}{15}\)
\(=\dfrac{-3}{5}+\dfrac{4}{9}-\dfrac{16}{15}\)
\(=\dfrac{\left(-27\right)+20-48}{45}=\dfrac{-55}{45}=\dfrac{-11}{9}\)
c) \(-\dfrac{15}{4}.\left(\dfrac{-7}{15}\right).\left(-2\dfrac{2}{5}\right)\)
\(=\dfrac{7}{4}.\dfrac{-12}{5}\)
\(=\dfrac{-21}{5}\)
\(#Wendy.Dang\)
1) <=>\(\left[\begin{array}{nghiempt}7-x=5x+1\\7-x=-5x-1\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=\frac{4}{3}\\x=-2\end{array}\right.\)
2)<=>\(\left[\begin{array}{nghiempt}x+1=3x+2\\x+1=-3x-2\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=-\frac{3}{2}\\x=-\frac{1}{4}\end{array}\right.\)
3)<=> \(\left[\begin{array}{nghiempt}x+15=3x-1\\x+15=1-3x\end{array}\right.\)
<=>\(\left[\begin{array}{nghiempt}x=8\\x=-\frac{7}{2}\end{array}\right.\)
4) \(\left[\begin{array}{nghiempt}x+7=x+7\\x+7=-x-7\end{array}\right.\)=> \(\left[\begin{array}{nghiempt}x\in R\\x=0\end{array}\right.\)
=> S=R
\(\frac{x+7}{x-3}=\frac{4}{15}\)
\(\left(x+7\right)\cdot15=\left(x-3\right)\cdot4\)
\(15x+105=4x-12\)
\(11x=-117\)
\(x=-\frac{117}{11}\)