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x4+4x3-4x2-48x-48=0
=> x4+4(x3-x2) - 48x = 48
=> x4 + 4[x2(x-1)] - 48x = 48
\(x^4+4x^3-4x^2-48x-48=0\)
\(\Leftrightarrow\)\(x^4-2x^3-4x^2+6x^3-12x^2-24x+12x^2-24x-48=0\)
\(\Leftrightarrow\)\(x^2\left(x^2-2x-4\right)+6x\left(x^2-2x-4\right)+12\left(x^2-2x-4\right)=0\)
\(\Leftrightarrow\)\(\left(x^2-2x-4\right)\left(x^2+6x+12\right)\)
\(\Leftrightarrow\)\(\left[\left(x-1\right)^2-5\right]\left(x^2+6x+12\right)=0\)
\(\Leftrightarrow\)\(\left(x-1-\sqrt{5}\right)\left(x-1+\sqrt{5}\right)\left(x^2+6x+12\right)=0\)
Ta có: \(x^2+6x+12=\left(x+3\right)^2+3>0\)
\(\Rightarrow\)\(\orbr{\begin{cases}x-1-\sqrt{5}=0\\x-1+\sqrt{5}=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1+\sqrt{5}\\x=1-\sqrt{5}\end{cases}}\)
Vậy...
\(\Delta=b^2-4ac=\left(-48\right)^2-4.1.\left(-25\right)=2400>0\)
do đó pt có 2 nghiệm phân biệt là:
\(•x_1=\dfrac{-b-\sqrt{\Delta}}{2a}=\dfrac{48-\sqrt{2400}}{2}=24-10\sqrt{6}\\ •x_2=\dfrac{-b+\sqrt{\Delta}}{2a}=\dfrac{48+\sqrt{2400}}{2}=24+10\sqrt{6}\)
\(x^2-48x-25=0\)
\(\Leftrightarrow x^2-2.x.24+24^2-601=0\)
\(\Leftrightarrow\left(x-24\right)^2-601=0\)
\(\Leftrightarrow\left(x-24\right)^2=601\)
\(\Leftrightarrow x-24=\sqrt{601}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-24=\sqrt{601}\\x-24=-\sqrt{601}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=24+\sqrt{601}\\x=24-\sqrt{601}\end{matrix}\right.\)
a, \(49x^2-70x+25=\left(7x\right)^2-2.7x.5+5^2=\left(7x-5\right)^2\)
Thay x = 5 vào biểu thức trên : \(\left(35-5\right)^2=30^2=900\)
b, \(x^3+12x^2+48x+64=\left(x+4\right)^3\)
Thay x = 6 vào biểu thức trên ta được : \(\left(6+4\right)^3=1000000\)
3, \(4x^2+4xy+y^2=\left(2x+y\right)^2\)
Thay x = -6 ; y = 2 vào biểu thức trên ta được : \(\left(-12+2\right)^2=100\)
a) 49x2 - 70x + 25 = (7x)2 - 2.7.5x + 52 = (7x - 5)2 = (7.5 - 5)2 = 302 = 900
b) x3 + 12x2 + 48x + 64 = (x + 4)3 = (6 + 4)3 = 103 = 1000
c) 4x2 + 4xy + y2 = (2x + y)2 = (-6.2 + 2)2 = (-10)2 = 100
d) x3 - 6x2 + 12x - 8 = (x - 2)3 = (102 - 2)3 = 1003 = 1000000
\(\Leftrightarrow\left(x+3\right)^2-48x^2=\left(x-3\right)^2\)
\(\Leftrightarrow48x^2=x^2+6x+9-x^2+6x-9\)
\(\Leftrightarrow48x^2-12x=0\)
=>12x(4x-1)=0
=>x=0(nhận) hoặc x=1/4(nhận)
\(=>A=x^3+3.4x^2+3.4^2x+4^3=\left(x+4\right)^3\)
tại x=6 \(=>A=\left(6+4\right)^3=10^3=1000\)
\(x^3+12x^2+48x+64\)
= \(\left(x^3+64\right)+\left(12x^2+48x\right)\)
= \(\left(x+4\right)\left(x^2-4x+16\right)+12x\left(x+4\right)\)
= \(\left(x+4\right)\left(x^2-4x+16+12x\right)\)
= \(\left(x+4\right)\left(x^2-8x+16\right)\)
= \(\left(x+4\right)\left(x-4\right)^2\)
Ta có: \(x^3+12x^2+48x+64=8x^3-12x^2+6x-1\)
\(\Leftrightarrow\left(x+2\right)^3=\left(2x-1\right)^3\)
\(\Leftrightarrow\left(x+2\right)^3-\left(2x-1\right)^3=0\)
\(\Leftrightarrow\left[\left(x+2\right)-\left(2x-1\right)\right]\left[\left(x+2\right)^2+\left(x+2\right)\left(2x-1\right)+\left(2x-1\right)^2\right]=0\)
\(\Leftrightarrow\left(x+2-2x+1\right)\left(x^2+4x+4+2x^2+3x-2+4x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(3-x\right)\left(7x^2+3x-6\right)=0\)
\(\Leftrightarrow7\left(3-x\right)\cdot\left(x^2+\frac{3}{7}x-\frac{6}{7}\right)=0\)
mà 7>0
nên \(\left[{}\begin{matrix}3-x=0\\x^2+\frac{3}{7}x-\frac{6}{7}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2+2\cdot x\cdot\frac{3}{14}+\frac{9}{196}-\frac{177}{196}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\\left(x+\frac{3}{14}\right)^2=\frac{177}{196}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+\frac{3}{14}=\frac{\sqrt{177}}{14}\\x+\frac{3}{14}=-\frac{\sqrt{177}}{14}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-3+\sqrt{177}}{14}\\x=\frac{-3-\sqrt{177}}{14}\end{matrix}\right.\)
Vậy: \(S=\left\{3;\frac{-3+\sqrt{177}}{14};\frac{-3-\sqrt{177}}{14}\right\}\)
\(48\left(x-2\right)=48x+25\)
\(\Rightarrow48x-48.2=48x+25\)
\(\Rightarrow48x-96=48x+25\)
\(\Rightarrow48x-48x=25+96=121\)
\(\Rightarrow0=121\)
=> Vô lí