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a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
\(a,\Leftrightarrow\left(x-9\right)^2-2\left(x-9\right)+1=0\\ \Leftrightarrow\left(x-9-1\right)^2=0\Leftrightarrow x=10\\ b,Sửa:49x^2-14x\sqrt{5}+5=0\\ \Leftrightarrow\left(7x-\sqrt{5}\right)^2=0\Leftrightarrow x=\dfrac{\sqrt{5}}{7}\)
A) \(\left(x+y\right)^2=\left(x-y\right)^2+4xy=5^2+4.3=37\)
B)
a) \(\left(x+3\right)^2-\left(x-2\right)^2=11\)
\(\Leftrightarrow\)\(x^2+6x+9-\left(x^2-4x+4\right)-11=0\)
\(\Leftrightarrow\)\(x^2+6x+9-x^2+4x-4-11=0\)
\(\Leftrightarrow\)\(10x-6=0\)
\(\Leftrightarrow\)\(10x=6\)
\(\Leftrightarrow\)\(x=\frac{3}{5}\)
Vậy...
b) \(25x^2-9=0\)
\(\Leftrightarrow\)\(\left(5x-3\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}\)
Vậy...
\(\left(2x+1\right)2-4\left(x+2\right)2=9\)
\(4x+2-8x-16=9\)
\(4x-8x=9+16-2\)
\(-4x=23\)
\(x=-\frac{23}{4}\)
a, \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x+4\right)\left(x-4\right)=5\)
\(\Rightarrow x^3+3x^2+9x-3x^2-9x-27-x\left(x^2-16\right)=5\)
\(\Rightarrow x^3-27-x^3-16x=5\)
\(\Rightarrow-16x-27=5\)
\(\Rightarrow-16x=32\Rightarrow x=-2\)
b, \(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)
\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-5x^2+25x+5x^2-25x+125\right)+6x^2=11\)
\(\Rightarrow x^3-6x^2+12x-8-x^3-125+6x^2=11\)
\(\Rightarrow12x-133=11\Rightarrow12x=144\Rightarrow x=12\)
Chúc bạn học tốt!!!
a)
\(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x+4\right)\left(x-4\right)=5\)
\(\Rightarrow x^3-3^3-x.\left(x^2-16\right)=5\)
\(\Rightarrow x^3-27-x^3+16.x=5\)
\(\Rightarrow16x-27=5\)
\(\Rightarrow16x=32\)
\(\Rightarrow x=2\)
Vậy x = 2
b)
\(\left(x-2\right)^3-\left(x+5\right)\left(x^2-5x+25\right)+6x^2=11\)
\(\Rightarrow x^3-6x^2+12x-8-x^3-125+6x^2=11\)
\(\Rightarrow12x-133=11\)
\(\Rightarrow12x=144\)
\(\Rightarrow x=12\)
Vậy x = 12
a,thay P(1),P(2),P(3),P(4) vào P(x(=) rồi giải hệ pt
câu b thì thay x=567 vào P(x) tính đc ở trên nhờ có các hệ số a,b,c,d
\(\left(x+3\right)\left(x-1\right)-x\) \(\left(x-5\right)=11\)
\(x^2-x+3x-3\) \(-x^2+5x=11\)
\(7x-3=11\)
\(7x=14\)
\(x=2\)
P \(=\left(x+3y\right)\) \(\left(x^2-3xy+9^2\right)\)
\(=\left(\frac{1}{2}+3.\frac{1}{2}\right)\) \(\left(\frac{1}{4}-3.\frac{1}{2}.\frac{1}{2}+81\right)\)
\(=2.\frac{161}{2}\)
\(=161\)
x=6 ok em
pt <=> 5.|x-2| = 9+ 11 = 20
<=> |x-2| = 20 : 5 =4
<=> x-2 = 4 hoặc x-2 = -4
<=> x=6 hoặc x= -2