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Đặt Q(x)=P(x)-10x. Khi đó Q(1)=Q(2)=Q(3)=0
Vì vậy Q(x) chia hết cho (x-1)(x-2)(x-3). Q(x) là đa thức bậc 4 (do P(x) là đa thức bậc 4) nên Q(x)=(x-1)(x-2)(x-3)(x-r) và
P(x)=(x-1)(x-2)(x-3)(x-r)+10x
P(12)=1200-990r
P(-8)=7840+990r
Vậy \(\frac{P\left(12\right)+P\left(-8\right)}{10}=1984\)
Ta có: \(P\left(1\right)=1+a+b+c+d=10\)
\(P\left(2\right)=16+8a+4b+2c+d=20\)
\(P\left(3\right)=81+27a+9b+3c+d=30\)
và \(P\left(12\right)=20736+1728a+144b+12c+d\)
\(P\left(-8\right)=4096-512a+64b-8c+d\)
suy ra \(P\left(12\right)+P\left(-8\right)=24832+1216a+208b+4c+2d\)
Ta lại có: \(100.P\left(1\right)-198.P\left(2\right)+100.P\left(3\right)\) \(=100\left(1+a+b+c+d\right)-198\left(16+8a+4b+2c+d\right)+100\left(81+27a+9b+3c+d\right)\)
\(=100+100a+100b+100c+100d-3168-1584a-792b-396c-198d+8100+2700a+900b+300c+100d\)
\(=5032+1216a+208b+4c+2d\)
Mặt khác: \(100.P\left(1\right)-198.P\left(2\right)+100.P\left(3\right)\)
\(=100\times10-198\times20+100\times30=40\)
Do đó: \(5032+1216a+208b+4c+2d=40\)
\(\Rightarrow\)\(1216a+208b+4c+2d=40-5032=-4992\)
Thế \(1216a+208b+4c+2d=-4992\) vào \(P\left(12\right)+P\left(-8\right)=24832+1216a+208b+4c+2d\)
ta được: \(P\left(12\right)+P\left(-8\right)=24832-4992=19840\)
Vậy \(\frac{P\left(12\right)+P\left(-8\right)}{10}=\frac{19840}{10}=1984\)
Lời giải:
Ta có thể viết dạng của $f(x)$ như sau:
\(f(x)=(x-1)(x-2)(x-3)(x-t)+g(x)\)
Trong đó, \(t\) là một số bất kỳ nào đó và \(g(x)\) là đa thức có bậc nhỏ hơn hoặc bằng $3$
Giả sử \(g(x)=mx^3+nx^2+px\)
\(\left\{\begin{matrix} f(1)=g(1)=m+n+p=10\\ f(2)=g(2)=8m+4n+2p=20\\ f(3)=g(3)=27m+9n+3p=30\end{matrix}\right.\)
Giải hệ trên thu được \(m=0,n=0,p=10\)
Như vậy \(f(x)=(x-1)(x-2)(x-3)(x-t)+10x\)
Do đó \(\left\{\begin{matrix} f(12)=990(12-t)+120=12000-990t\\ f(-8)=-990(-8-t)-80=7840+990t\end{matrix}\right.\)
\(\Rightarrow \frac{f(12)+f(-8)}{10}+26=\frac{12000+7840}{10}+26=2010\) (đpcm)
Có lẽ bạn nên sửa đề thành \(f\left(x\right)=...x^2+1...\)hoặc là \(g\left(x\right)=...\left(bx-1\right)...\)
Ta có:
\(f\left(x\right)=ax^3+4x^3-4x+8=\left(a+4\right)x^3-4x+8\)
\(g\left(x\right)=x^3+4x\left(bx-1\right)+c-3=x^3+4bx^2-4x+c-3\)
Để \(f\left(x\right)=g\left(x\right)\Leftrightarrow\hept{\begin{cases}a+4=1\\4b=0\\c-3=8\end{cases}\Leftrightarrow\hept{\begin{cases}a=-3\\b=0\\c=11\end{cases}}}\)
Kết luận
\(a,-\frac{3}{2}-2x+\frac{3}{4}=-2\)
=> \(-\frac{3}{2}+\left(-2x\right)+\frac{3}{4}=-2\)
=> \(\left(-\frac{3}{2}+\frac{3}{4}\right)+\left(-2x\right)=-2\)
=> \(-\frac{3}{4}+\left(-2x\right)=-2\)
=> \(-2x=-2-\left(-\frac{3}{4}\right)=-\frac{5}{4}\)
=> \(x=-\frac{5}{4}:\left(-2\right)=\frac{5}{8}\)
Vậy \(x\in\left\{\frac{5}{8}\right\}\)
\(b,\left(\frac{-2}{3}x-\frac{3}{4}\right)\left(\frac{3}{-2}-\frac{10}{4}\right)=\frac{2}{5}\)
=> \(\left(-\frac{2}{3}x-\frac{3}{4}\right).\left(-4\right)=\frac{2}{5}\)
=> \(-\frac{2}{3}x-\frac{3}{4}=\frac{2}{5}:\left(-4\right)=-\frac{1}{10}\)
=> \(-\frac{2}{3}x=-\frac{1}{10}+\frac{3}{4}=\frac{13}{20}\)
=> \(x=\frac{13}{20}:\left(-\frac{2}{3}\right)=-\frac{39}{40}\)
Vậy \(x\in\left\{-\frac{39}{40}\right\}\)
\(c,\frac{x}{2}-\left(\frac{3x}{5}-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
=> \(\frac{x}{2}-\frac{3x}{5}+\frac{13}{5}=-\frac{7}{5}-\frac{7}{10}x\)
=> \(10.\frac{x}{2}-10.\frac{3x}{5}+10.\frac{13}{5}=10.\frac{-7}{5}-10.\frac{7}{10}x\)
( chiệt tiêu )
=> \(5x-6x+26=-14-7x\)
=> \(-x+26=-14-7x\)
=> \(-x+7x=-14-26\)
=> \(6x=-40\)
=> \(x=-40:6=\frac{20}{3}\)
Vậy \(x\in\left\{\frac{20}{3}\right\}\)
\(d,\frac{2x-3}{3}+\frac{-3}{2}=\frac{5-3x}{6}-\frac{1}{3}\)
=> \(6.\frac{2x-3}{3}+6.\frac{-3}{2}=6.\frac{5-3x}{6}-6.\frac{1}{3}\)
( chiệt tiêu )
=> \(2\left(2x-3\right)-9=5-3x-2\)
=> \(4x-6-9=3-3x\)
=> \(4x-15=3-3x\)
=> \(4x+3x=3+15\)
=> \(7x=18\)
=> \(x=18:7=\frac{18}{7}\)
Vậy \(x\in\left\{\frac{18}{7}\right\}\)
\(e,\frac{2}{3x}-\frac{3}{12}=\frac{4}{x}-\left(\frac{7}{x}.2\right)\)
ĐKXĐ : \(x\ne0\)
=> \(\frac{2}{3x}-\frac{1}{4}=\frac{4}{x}-\frac{14}{x}\)
=> \(\frac{2}{3x}-\frac{4}{x}+\frac{14}{x}=\frac{1}{4}\)
=> \(\frac{2}{3x}-\frac{12}{3x}+\frac{42}{3x}=\frac{1}{4}\)
=> \(\frac{32}{3x}=\frac{1}{4}\)
=> \(3x=32.4:1=128\)
=> \(x=128:3=\frac{128}{3}\)
Vậy \(x\in\left\{\frac{128}{3}\right\}\)
\(k,\frac{13}{x-1}+\frac{5}{2x-2}-\frac{6}{3x-3}\)
ĐKXĐ :\(x\ne1;\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{6}{3\left(x-1\right)}\)
=> \(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}-\frac{1}{x-1}\)
=> \(\frac{2.13}{2\left(x-1\right)}+\frac{5}{2\left(x-1\right)}-\frac{2.1}{2.\left(x-1\right)}\)
=> \(\frac{26+5-2}{2\left(x-1\right)}\)
=> \(\frac{29}{2\left(x-1\right)}\)
\(m,\left(\frac{3}{2}-\frac{2}{-5}\right):x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x-\frac{1}{2}=\frac{3}{2}\)
=> \(\frac{19}{10}:x=\frac{3}{2}+\frac{1}{2}=2\)
=> \(x=\frac{19}{10}:2=\frac{19}{20}\)
Vậy \(x\in\left\{\frac{19}{20}\right\}\)
\(n,\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\left(2x-1\right)=\left(\frac{-3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)
=> \(\frac{233}{286}\left(2x-1\right)=-\frac{233}{572}\)
=> \(2x-1=-\frac{233}{572}:\frac{233}{286}=-\frac{1}{2}\)
=> \(2x=-\frac{1}{2}+1=\frac{1}{2}\)
=> \(x=\frac{1}{2}:2=\frac{1}{4}\)
Vậy \(x\in\left\{\frac{1}{4}\right\}\)
a) \(\left(2x-1\right)^3=-8\)
\(\left(2x-1\right)^3=\left(-2\right)^3\)
=> 2x - 1 = -2
=> x = -1/2
a) \(-6.\left(-\frac{2}{3}\right).0,25\)
\(=-6.\left(-\frac{2}{3}\right).\frac{1}{4}\)
\(=4.\frac{1}{4}\)
\(=1.\)
b) \(-\frac{15}{4}.\left(-\frac{7}{15}\right).\left(-2\frac{2}{5}\right)\)
\(=-\frac{15}{4}.\left(-\frac{7}{15}\right).\left(-\frac{12}{5}\right)\)
\(=\frac{7}{4}.\left(-\frac{12}{5}\right)\)
\(=-\frac{21}{5}.\)
c) \(\left(-0,4\right)^2-\left(-0,4\right)^3.\left(-3\right)\)
\(=0,16-\left(-0,064\right).\left(-3\right)\)
\(=0,16-0,192\)
\(=-0,032.\)
Chúc bạn học tốt!
\(P\left(x\right)=x^4+ax^3+bx^2+cx+d\)
Đặt \(Q\left(x\right)=P\left(x\right)-10x\) \(\Rightarrow\left\{{}\begin{matrix}Q\left(1\right)=P\left(1\right)-10.1=10-10=0\\Q\left(2\right)=P\left(2\right)-10.2=20-20=0\\Q\left(3\right)=P\left(3\right)-10.3=30-30=0\end{matrix}\right.\)
\(\Rightarrow Q\left(x\right)\) có 3 nghiệm \(x=\left\{1;2;3\right\}\Rightarrow Q\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-a\right)\)
Mà \(Q\left(x\right)=P\left(x\right)-10x\Rightarrow P\left(x\right)=Q\left(x\right)+10x\)
\(\Rightarrow P\left(x\right)=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-a\right)+10x\)
\(P\left(12\right)=990\left(12-a\right)+120=12000-990a\)
\(P\left(-8\right)=-990\left(-8-a\right)-80=990a+7840\)
\(\Rightarrow\frac{P\left(12\right)+P\left(-8\right)}{10}=\frac{12000-990a+990a+7840}{10}=1984\)