bài 4; tìm x
e, ( x mũ 3 - 4x mũ 2 ) - ( x - 4 ) = 0
f, 2x mũ 3 - 242x = 0
g., x mũ 5 - 9x = 0
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a, \(5x\left(x-1\right)+\left(x+17\right)=0\)
\(\Leftrightarrow5x^2-5x+x+17=0\Leftrightarrow5x^2-4x+17=0\)
\(\Leftrightarrow5\left(x^2-\frac{4}{5}x\right)+17=0\Leftrightarrow5\left(x^2-2.\frac{2}{5}x+\frac{4}{25}-\frac{4}{25}\right)+17=0\)
\(\Leftrightarrow5\left(x-\frac{2}{5}\right)^2-\frac{4}{5}+17=0\Leftrightarrow5\left(x-\frac{2}{5}\right)^2+81\ge81>0\)
Vậy pt vô nghiệm
b, \(3x\left(x-3\right)^2-3x\left(x+3\right)^2=0\)
\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)
\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\Leftrightarrow x.2x=0\Leftrightarrow x=0\)
c, \(2x^2-9x+7=0\Leftrightarrow2x^2-7x-2x+7=0\)
\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\Leftrightarrow\left(x-1\right)\left(2x-7\right)=0\Leftrightarrow x=1;x=\frac{7}{2}\)
Trả lời:
a, \(5x\left(x-1\right)+\left(x+17\right)=0\)
\(\Leftrightarrow5x^2-5x+x+17=0\)
\(\Leftrightarrow5x^2-4x+17=0\)
\(\Leftrightarrow5\left(x^2-\frac{4}{5}x+\frac{17}{5}\right)=0\)
\(\Leftrightarrow x^2-\frac{4}{5}x+\frac{17}{5}=0\)
\(\Leftrightarrow x^2-2.x.\frac{2}{5}+\frac{4}{25}+\frac{81}{25}=0\)
\(\Leftrightarrow\left(x-\frac{2}{5}\right)^2+\frac{81}{25}=0\)
Vì \(\left(x-\frac{2}{5}\right)^2+\frac{81}{25}\ge\frac{81}{25}>0\forall x\)
nên pt vô nghiệm
b, \(3x\left(x-3\right)^2-3x\left(x+3\right)^2=0\)
\(\Leftrightarrow3x\left[\left(x-3\right)^2-\left(x+3\right)^2\right]=0\)
\(\Leftrightarrow3x\left(x-3-x-3\right)\left(x-3+x+3\right)=0\)
\(\Leftrightarrow3x.\left(-9\right).2x=0\)
\(\Leftrightarrow-54x^2=0\)
\(\Leftrightarrow x^2=0\)
\(\Leftrightarrow x=0\)
Vậy x = 0 là nghiệm của pt.
c, \(7-9x+2x^2=0\)
\(\Leftrightarrow2x^2-7x-2x+7=0\)
\(\Leftrightarrow x\left(2x-7\right)-\left(2x-7\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=1\end{cases}}}\)
Vậy x = 7/2; x = 1 là nghiệm của pt.
d, trùng ý c
,(3x-1) mũ 2=9/16
<=> (3x-1)^2 = ( ±3/4)^2
<=> l3x-1l = 3/4
Hoặc 3x-1 = 3/4
<=> 3x= 3/4 + 1
<=> x = 7/4 : 3
<=> x= 7/1
Ta có:
A=310.11+310.539.24=310.(11+5)39.16A=310.11+310.539.24=310.(11+5)39.16
=310.1639.16=3.11.1=3=310.1639.16=3.11.1=3
Vậy giá trị biểu thức A là 3
s) = ( x2 - 2xy + y2 ) - ( 2xy )2 = ( x - y - 2xy )( x - y + 2xy )
u) sửa +4y thành -4y
= 4( x - y ) - x2( x - y ) = ( x - y )( 2 - x )( 2 + x )
a)x²−2x−4y²−4ya)x²-2x-4y²-4y
=x²−2x−4y²−4y+2xy−2xy=x²-2x-4y²-4y+2xy-2xy
=(x²−2xy−2x)+(2xy−4y²−4y)=(x²-2xy-2x)+(2xy-4y²-4y)
=x(x−2y−2)+2y(x−2y−2)=x(x-2y-2)+2y(x-2y-2)
=(x+2y)(x−2y−2)=(x+2y)(x-2y-2)
b)x4+2x³−4x−4b)x4+2x³-4x-4
=x4+2x³+2x²−2x²−4x−4=x4+2x³+2x²-2x²-4x-4
=(x4+2x³+2x²)−(2x²+4x+4)=(x4+2x³+2x²)-(2x²+4x+4)
=x²(x²+2x+2)−2(x²+2x+2)=x²(x²+2x+2)-2(x²+2x+2)
=(x²−2)(x²+2x+2)=(x²-2)(x²+2x+2)
c)x³+2x²y−x−2yc)x³+2x²y-x-2y
=x²(x+2y)−(x+2y)=x²(x+2y)-(x+2y)
=(x²−1)(x+2y)=(x²-1)(x+2y)
=(x+1)(x−1)(x+2y)=(x+1)(x-1)(x+2y)
d)3x²−3y²−2(x−y)²d)3x²-3y²-2(x-y)²
=3(x²−y²)−2(x−y)²=3(x²-y²)-2(x-y)²
=3(x+y)(x−y)−2(x−y)²=3(x+y)(x-y)-2(x-y)²
=(x−y)[3(x+y)−2(x−y)]=(x-y)[3(x+y)-2(x-y)]
=(x−y)(3x+3y−2x+2y)=(x-y)(3x+3y-2x+2y)
=(x−y)(x+5y)=(x-y)(x+5y)
e)x³−4x²−9x+36e)x³-4x²-9x+36
=(x³−4x²)−(9x−36)=(x³-4x²)-(9x-36)
=x²(x−4)−9(x−4)=x²(x-4)-9(x-4)
=(x−4)(x²−9)=(x-4)(x²-9)
=(x−4)(x²−3²)=(x-4)(x²-3²)
=(x−4)(x+3)(x−3)=(x-4)(x+3)(x-3)
f)x²−y²−2x−2yf)x²-y²-2x-2y
=(x²−y²)−(2x+2y)=(x²-y²)-(2x+2y)
=(x+y)(x−y)−2(x+y)=(x+y)(x-y)-2(x+y)
=(x+y)(x−y−2)
hok tốt nhé
k đi
đm con chó
e, \(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-4\right)-\left(x-4\right)=0\Leftrightarrow\left(x^2-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x-4\right)=0\Leftrightarrow x=\pm1;x=4\)
f, \(2x^3-242x=0\Leftrightarrow2x\left(x^2-121\right)=0\)
\(\Leftrightarrow2x\left(x-11\right)\left(x+11\right)=0\Leftrightarrow x=\pm11;x=0\)
g, \(x^5-9x=0\Leftrightarrow x\left(x^4-9\right)=0\)
\(\Leftrightarrow x\left(x^2-3\right)\left(x^2+3>0\right)=0\Leftrightarrow x=\pm\sqrt{3};x=0\)