tim x biet : x^2+7x+10=0
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a) 7x=14
x = 14 : 7
x = 2
b ) 6x - (-5) = 17
6x = 17+(-5)
6x = 12
x = 12 :6
x = 2
c) (x+2) ( x-9) = 0
=> x +2 = 0 hoặc x-9=0
=> x= -2 => x=9
mình làm rùi , đi
Ta có:
x2 + 7x + 10 = 0
<=> x^2 + 5x + 2x + 10 = 0
<=> x(x + 5) + 2(x + 5) = 0
<=> (x+2)(x+5) = 0
<=> x+2=0 hoặc x+5=0
<=> x= -2 hoặc x= -5
Vậy x = -2; -5.
\(x^2+7x+10=0\)
\(\Leftrightarrow\left(x^2+5x\right)+\left(2x+10\right)=0\)
\(\Leftrightarrow x\left(x+5\right)+2\left(x+5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x+5=0\\x+2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-5\\x=-2\end{cases}}}\)
\(b,4x^2-x-5=0\)
\(\Leftrightarrow4x^2-5x+4x-5=0\)
\(\Leftrightarrow x\left(4x-5\right)+4x-5=0\)
\(\Leftrightarrow\left(4x-5\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{4}\end{cases}}\)
Bài 2
\(a,x^3+5x^2+3x-9\)
\(\Leftrightarrow x^3-x^2+6x^2-6x+9x-9\)
\(\Leftrightarrow x^2\left(x-1\right)+6x\left(x-1\right)+9\left(x-1\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+6x+9\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)^2\)
b,\(x^3-7x-6\)
\(\Leftrightarrow x^3-3x^2+3x^2-9x+2x-6\)
\(\Leftrightarrow x^2\left(x-3\right)+3x\left(x-3\right)+2\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+2\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)\left(x+2\right)\)
c,\(3x^3-7x^2+17x-5\)
\(\Leftrightarrow3x^3-x^2-6x^2+2x+15x-5\)
\(\Leftrightarrow x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-2x+5\right)\)
Bài 1:
a)\(28x^3+15x^2+75x+125=0\)
\(\Leftrightarrow\left(4x+5\right)\left(7x^2-5x+25\right)=0\)
Dễ thấy: \(7x^2-5x+25=7\left(x-\frac{5}{14}\right)^2+\frac{675}{28}>0\)
\(\Rightarrow4x+5=0\Rightarrow x=-\frac{5}{4}\)
b)\(4x^2-x-5=0\)
\(\Leftrightarrow\left(x+1\right)\left(4x-5\right)=0\)
\(\Rightarrow x=-1;x=\frac{5}{4}\)
Bài 2:
a)\(x^3+5x^2+3x-9\)
\(=\left(x-1\right)\left(x+3\right)^2\)
b)\(x^3-7x-6\)
\(=\left(x-3\right)\left(x+1\right)\left(x+2\right)\)
c)\(3x^3-7x^2+17x-5\)
\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
\(3x+4=0\Leftrightarrow x=-\dfrac{4}{3}\\ 2x\left(x-1\right)-\left(1+2x\right)=-34\\ \Leftrightarrow2x^2-2x-1-2x=-34\\ \Leftrightarrow2x^2-4x+33=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+30=0\\ \Leftrightarrow2\left(x-1\right)^2+30=0\\ \Leftrightarrow x\in\varnothing\left[2\left(x-1\right)^2+30\ge30>0\right]\\ x^2+9x-10=0\\ \Leftrightarrow x^2-x+10x-10=0\\ \Leftrightarrow\left(x-1\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-10\end{matrix}\right.\\ \left(7x-1\right)\left(2+5x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}7x-1=0\\2+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{7}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
(7x - 11)3 = 25 * 22 + 200
= (7x - 11)3 = 128 + 200
= 7x - 113 = 328
= 7x - 1331 = 328
= 7x = 328 + 1331
= 7x = 1659
= x = 1659 : 7
= x = 237
Vậy x = 237
Do \(7x=3y\)\(\Rightarrow\frac{x}{3}=\frac{y}{7}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x}{3}=\frac{y}{7}=\frac{x-y}{3-7}=\frac{16}{-4}=-4\) ( do x - y = 16 )
Khi đó:
\(\frac{x}{3}=-4\)\(\Rightarrow x=\left(-4\right)\cdot3=-12\)
\(\frac{y}{7}=-4\)\(\Rightarrow y=\left(-4\right)\cdot7=-28\)
Vậy x = -12 ; y = -28
(X+2)(X+5)=0
=>x=-2 hoặc x=-5
x^2+7x+10=0
x(x+7)=-10
=>x>0 x<0
x+7<0 x+7<0
Mà x+7>x
=>x<0 =>x<0
x+7>0 x>-7
=>x thuộc -1;-2;-3;-4;-5;-6