4x^2-25-(2x-5)(2x+7)
4x^2-25 - (2x-5)* (2x+7)
\(4x^2-25-\left(2x-5\right)\left(2x+7\right)\)
\(=4x^2-25-\left(4x^2+4x-35\right)\)
\(=4x^2-25-4x^2-4x+35\)
\(=35-25-4x\)
\(=10-4x\)
\(=2\left(5-2x\right)\)
4x^2- 25-(2x-5)(2x-7)=0
4x^2-25-(2x-5)(2x+7)=0
<=> 4x^2 - 25 - (4x^2 + 14x - 10x - 35) = 0
<=> 4x^2 - 25 - 4x^2 - 14x + 10x + 35 = 0
<=> -4x + 10 = 0
<=> x = -10/-4
<=> x = 5/2
4x^2-25-(2x-5)(2x+7)=0
<=> 4x^2 - 25 - (4x^2 + 14x - 10x - 35) = 0
<=> 4x^2 - 25 - 4x^2 - 14x + 10x + 35 = 0
<=> -4x + 10 = 0
<=> x = -10/-4
<=> x = 5/2
4x2 - 25 - (2x - 5)( 2x + 7) = 0
\(4x^2-25-\left(2x-5\right)\left(2x+7\right)=0\\ \Leftrightarrow4x^2-25-\left(4x^2+14x-10x-35\right)=0\\ \Leftrightarrow4x^2-25-4x^2-14x+10x+35=0\\
\Leftrightarrow-4x+10=0\)
\(\Leftrightarrow x=\frac{-10}{-4}\\
\Leftrightarrow x=\frac{5}{2}\)
1) 4x2 – 25 + (2x + 7)(5 – 2x)
Ta có đa thức tương đương: 4x2-25+10x-4x2+35-14x
=-4x+10=-2(2x+5)
tìm x :
(5-2x)x(2x+7)=4x^2-25
\(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)=\left(2x+5\right)\left(2x-5\right)\)
\(\Leftrightarrow-\left(2x-5\right).\left(2x+7\right)-\left(2x+5\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(-7-2x-2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(-12-4x\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\-12-4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)
Vậy x=5/2 hoặc x=-3
Tìm x biết : \(\left(5-2x\right)\left(2x+7\right)-4x^2-25=0\)
\(\left(5x-2\right)\left(2x+7\right)-4x^2-25=0\)
\(10x+35-4x^2-14x-4x^2+25=0\)
\(-4x+60-8x^2=0\)
\(-4\left(2x^2+x-15\right)=0\)
\(-4\left(2x^2+6x-5x-15\right)=0\)
\(-4\left(2x-5\right)\left(x+3\right)=0\)
=> \(x\) ∈ \(\left\{\dfrac{5}{2};-3\right\}\)
4x2 - 25 + (2x+7)(5-2x)
\(4x^2-25+\left(2x+7\right)\left(5-2x\right)\)
\(=4x^2-25+10x-4x^2+35-14x\)
\(=-4x+10=2\left(-2x+5\right)\)
tim X \(4x^2-25=(2x-5)(2x+7)\)
\(4x^2-25=\left(2x-5\right).\left(2x+7\right)\)
\(\Rightarrow\left(2x\right)^2-5^2=\left(2x-5\right).\left(2x+7\right)\)
\(\Rightarrow\left(2x-5\right).\left(2x+5\right)-\left(2x-5\right).\left(2x+7\right)=0\)
\(\Rightarrow\left(2x-5\right).\left(2x+5-2x-7\right)=0\)
\(\Rightarrow\left(2x-5\right).\left(-2\right)=0\)
Vì \(-2\ne0.\)
\(\Rightarrow2x-5=0\)
\(\Rightarrow2x=0+5\)
\(\Rightarrow2x=5\)
\(\Rightarrow x=5:2\)
\(\Rightarrow x=\frac{5}{2}\)
Vậy \(x=\frac{5}{2}.\)
Chúc bạn học tốt!
4x2−25=(2x−5)(2x+7)
4x2−25 = 4x2 + 14x -10x - 35
4x2−25 - ( 4x2 + 14x -10x - 35) = 0
4x2 −25 - 4x2 -14x + 10x + 35 = 0
-4x + 10 = 0
=> x = -\(\frac{5}{2}\)
Vậy x = -\(\frac{5}{2}\)
Câu này bình thường mà bạn có thẻ giải bằng cách dùng hàng đẳng thức số 3 hoặc nhân tung vào như vậy cũng đc
Phân tích đa thức thành nhân tử: 4x^2 -25+(2x+7).(5-2x)
Help me,
\(=\left(2x-5\right)\left(2x+5\right)-\left(2x-5\right)\left(2x+7\right)\)
\(=\left(2x-5\right)\left(2x+5-2x-7\right)\)
\(=-2\left(2x-5\right)\)