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\(Q\left(x\right)-P\left(x\right)=0\)
\(\Leftrightarrow\left(-6x^2+x^3-8+12\right)-\left(x^3-3x^2+6x-8\right)=0\)
\(\Leftrightarrow\left(-6x^2+x^3+4\right)-\left(x^3-3x^2+6x-8\right)=0\)
\(\Leftrightarrow-6x^2+x^3+4-x^3+3x^2-6x+8=0\)
\(\Leftrightarrow-3x^2-6x+12=0\)
\(\Leftrightarrow-3\left(x^2+2x-4\right)=0\)
\(\Leftrightarrow x^2+2x-4=0\)
\(\Leftrightarrow x^2+2x+1=5\)
\(\Leftrightarrow\left(x+1\right)^2=5\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=\sqrt{5}\\x+1=-\sqrt{5}\end{cases}}\Leftrightarrow x=\pm\sqrt{5}-1\)
\(P\left(x\right)-Q\left(x\right)=\left(x^3-3x^2+6x-8\right)-\left(-6x^2+x^3-8+12\right)\)
\(P\left(x\right)-Q\left(x\right)=\left(x^3-3x^2+6x-8\right)-\left(-6x^2+x^3+4\right)\)
\(P\left(x\right)-Q\left(x\right)=x^3-3x^2+6x-8+6x^2-x^3-4\)
\(P\left(x\right)-Q\left(x\right)=3x^2+6x-4\)
Ta cần phân tích \(3x^2+6x-4\) thành nhân tử
Ta có:\(P\left(x\right)-Q\left(x\right)=-\frac{1}{3}\left(-9x^2-18x+12\right)\)
\(=-\frac{1}{3}\left[21-\left(9x^2+18x+9\right)\right]\)
\(=-\frac{1}{3}\left[21-\left(3x+3\right)^2\right]\)
\(=-\frac{1}{3}\left(\sqrt{21}-3x-3\right)\left(\sqrt{21}+3x+3\right)\)
\(\Rightarrow x=\frac{\sqrt{21}-3}{3};x=\frac{-\sqrt{21}-3}{3}\)
a) \(A=\frac{5^4.20^4}{25^5.4^5}=\frac{5^4.\left(2^2.5\right)^4}{5^{2^5}.\left(2^2\right)^5}=\frac{5^8.2^8}{5^{10}.2^{10}}=\frac{1}{\left(5^{10}:5^8\right).\left(2^{10}:2^8\right)}=\frac{1}{5^2.2^2}=\frac{1}{25.4}=\frac{1}{100}\)
b) \(B=\frac{2^{30}.5^7+2^{13}.5^{27}}{2^{27}.5^7+2^{10}.5^{27}}\)\(=\frac{2^3+2^3}{1}=\frac{8+8}{1}=16\)
c) \(C=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...........+\frac{1}{2^{100}}\)
\(\Rightarrow2C=1+\frac{1}{2}+\frac{1}{2^2}+..........+\frac{1}{2^{99}}\)
\(\Rightarrow2C-C=\left(1+\frac{1}{2}+\frac{1}{2^2}+.........+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...........+\frac{1}{2^{100}}\right)\)
\(\Rightarrow C=1-\frac{1}{2^{100}}\)
d) \(D=1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+.........+\frac{1}{5^{100}}\)
\(\Rightarrow5D=5+1+\frac{1}{5^2}+\frac{1}{5^3}+...........+\frac{1}{5^{101}}\)
\(\Rightarrow5D-D=\left(5+1+\frac{1}{5^2}+\frac{1}{5^3}+.........+\frac{1}{5^{101}}\right)-\left(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+..........+\frac{1}{5^{100}}\right)\)
\(\Rightarrow4D=5-\frac{1}{5^{101}}\)
\(\Rightarrow D=\frac{5-\frac{1}{5^{101}}}{4}\)
a) \(A=\frac{5^4x20^4}{25^5x4^5}=\frac{5^4x\left(2^2x5\right)^4}{\left(5^2\right)^5x\left(2^2\right)^5}=\frac{5^8.2^8}{5^{10}.2^{10}}=\frac{1}{5^2x2^2}=\frac{1}{25.4}=\frac{1}{100}\)
b) \(B=\frac{2^{30}x5^7+2^{13}x5^{27}}{2^{27}x5^7+2^{10}x5^{27}}=\frac{2^{13}.5^7.\left(2^{17}+5^{20}\right)}{2^{10}.5^7.\left(2^{17}+5^{20}\right)}=2^3=8\)
c) \(C=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2C=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2C-C=1-\frac{1}{2^{100}}\)
\(C=1-\frac{1}{2^{100}}\)
phần d bn lm tương tự như phần c nha!
a, 5n+5n+2=650
=>5n+5n.52=650
=>5n(1+25)=650
=>5n.26=650
=>5n=25
=>5n=52
=>n=2
Vậy n=2
1) Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{12x-15y}{7}=\frac{20y-12x}{9}=\frac{15y-20z}{11}=\frac{12x-15y+20z-12x+15y-20z}{7+9+11}=\frac{0}{27}=0\)
\(\Rightarrow\hept{\begin{cases}12x-15y=0\\15y-20z=0\end{cases}\Rightarrow}\hept{\begin{cases}12x=15y\\15y=20z\end{cases}\Rightarrow\hept{\begin{cases}\frac{x}{15}=\frac{y}{12}\\\frac{y}{20}=\frac{z}{15}\end{cases}\Rightarrow}\hept{\begin{cases}\frac{x}{75}=\frac{y}{60}\\\frac{y}{60}=\frac{z}{45}\end{cases}\Rightarrow}\frac{x}{75}=\frac{y}{60}=\frac{z}{45}}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x}{75}=\frac{y}{60}=\frac{z}{45}=\frac{x+y+z}{75+60+45}=\frac{48}{180}=\frac{4}{15}\)
=> x = 75.4 : 15 = 20 ;
y = 60.4 : 15 = 16 ;
z = 45.4 : 15 = 12
Vậy x = 20 ; y = 16 ; z = 12
2) Từ đẳng thức \(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)
\(\Rightarrow\frac{z}{y+z+t}+1=\frac{y}{z+t+x}+1=\frac{z}{t+x+y}+1=\frac{t}{x+y+z}+1\)
\(\Rightarrow\frac{x+y+z+t}{y+z+t}=\frac{x+y+z+t}{z+t+x}=\frac{x+y+z+t}{t+x+y}=\frac{x+y+z+t}{x+y+z}\)
Nếu x + y + z + t = 0
=> x + y = - (z + t)
=> y + z = - (t + x)
=> z + t = - (x + y)
=> t + x = - (z + y)
Khi đó :
P = \(\frac{-\left(z+t\right)}{z+t}+\frac{-\left(t+x\right)}{t+x}+\frac{-\left(x+y\right)}{x+y}+\frac{-\left(z+y\right)}{z+y}=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)=-4\)
=> P = 4
Nếu x + y + z + t khác 0
=> \(\frac{1}{y+z+t}=\frac{1}{z+t+x}=\frac{1}{t+x+y}=\frac{1}{x+y+z}\)
=> y + z + t = z + t + x = t + x + y = x + y + z
=> x =y = z = t
Khi đó : P = 1 + 1 + 1 + 1 = 4
Vậy nếu x + y + z + t = 0 thì P = - 4
nếu x + y + z + t khác 0 thì P = 4
a) \(\frac{3x+2}{-4x+5}=-\frac{4}{3}\left(ĐKXĐ:x\ne\frac{5}{4}\right)\)
\(\Rightarrow3\left(3x+2\right)=-4\left(-4x+5\right)\)
\(\Leftrightarrow9x+6=16x-20\)
\(\Leftrightarrow7x=26\)
\(\Leftrightarrow x=\frac{26}{7}\)
b) \(\frac{2\left|x\right|+5}{-4x+3}=-\frac{5}{4}\)(Thôi bài sau tự tìm đkxđ nhá)
\(\Rightarrow8\left|x\right|+20=20x-15\)
\(\Leftrightarrow8\left|x\right|-20x+35\)\(\left(1\right)\)
TH1: Nếu \(x\ge0\)thì \(\left(1\right)\Leftrightarrow8x-20x+35=0\Leftrightarrow x=\frac{35}{12}\left(tm\right)\)
TH2: Nếu \(x< 0\)thì \(\left(1\right)\Leftrightarrow-8x-20x+35=0\Leftrightarrow x=\frac{35}{28}\left(ktm\right)\)
Vậy x=35/12
c)\(\frac{2x+1}{5}=\frac{3}{2x-1}\)
\(\Rightarrow4x^2-1=15\)
\(\Leftrightarrow4x^2=16\)
\(\Leftrightarrow x^2=4\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
d)\(\frac{x+1}{2x+1}=\frac{0,5x+2}{x+3}\)
\(\Leftrightarrow\left(x+1\right)\left(x+3\right)=\left(2x+1\right)\left(0,5x+2\right)\)
\(\Leftrightarrow x^2+4x+3=x^2+4,5x+2\)
\(\Leftrightarrow0,5x=1\)
\(\Leftrightarrow x=2\)
e) \(\frac{\left|6x+1\right|}{4}=\frac{2}{4}\)
\(\Leftrightarrow\left|6x+1\right|=2\)
\(\Leftrightarrow\orbr{\begin{cases}6x+1=2\\6x+1=-2\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{6}\\x=-\frac{1}{2}\end{cases}}}\)
g)\(\frac{\left|3x-5\right|}{3}=\frac{\left|x\right|}{2}\)
\(\Leftrightarrow\frac{\left|3x-5\right|}{\left|x\right|}=\frac{3}{4}\)
\(\Leftrightarrow\left|\frac{3x-5}{x}\right|=\frac{3}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}\frac{3x-5}{x}=\frac{3}{4}\\\frac{3x-5}{x}=-\frac{3}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{20}{9}\\x=\frac{4}{3}\end{cases}}}\)
Mỏi tay quá, xin tý cho sảng khoái nào!!
\(\)
a) \(f\left(x\right)=-x^4+3x^3-\frac{1}{3}x^2+2x+5\)
\(g\left(x\right)=x^4+3x^3-\frac{2}{3}x^2-2x-10\)
b) \(f\left(x\right)+g\left(x\right)=-x^4+3x^3-\frac{1}{3}x^2+2x+5+x^4+3x^3-\frac{2}{3}x^2-2x-10\)
\(=6x^3-x^2-5\)
c) +) Thay x=1 vào đa thức f(x) + g(x) ta được :
\(6.1^3-1^2-5=0\)
Vậy x=1 là nghiệm của đa thức f(x) + g(x)
+) Thay x=-1 vào đa thức f(x) + g(x) ta được :
\(6.\left(-1\right)^3-\left(-1\right)^2-5=-10\)
Vậy x=-1 ko là nghiệm của đa thức f(x) + g(x)
b, x-2+3x =10 =>2.(2x-1)=2.5 =>4x-2=10 =>4x=10+2 =>4x=12 =>x=12:4 => x=3 Vậy x=3. Mk làm đại đúng thì đúng sai thì sai nha nhg mk đoán thì đúng
a)3x−1+5.3x−1=162
⇔6.3x−1=162
⇔3x−1=27
⇔3x−1=33
⇔x−1=3
⇔x=4