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TN
1
TN
24 tháng 7 2016
a) 4x^2 - 25 - ( 2x - 5) .( 2x + 7) = 0
<=>4x2-25-(4x2+14x-10x-35)=0
<=>4x2-25-4x2-14x+10x+35= 0
<=>-4x+10= 0
<=>x= 5/2
b) x^3 + 27 + ( x+3). ( x -9) = 0
<=>x3+33+(x+3)(x-9)=0
<=>(x+3)(x2-3x+9)+(x+3)(x-9)=0
<=>(x+3)(x2-3x+9+x-9) =0
<=>(x+3)(x2-2x)=0
<=>(x+3)(x-2)x= 0
<=>x=-3 hoặc x=2 hoặc x=2
27 tháng 6 2017
\(\left(5-2x\right)\left(2x+7\right)=4x^2-25\)\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)=\left(2x+5\right)\left(2x-5\right)\)
\(\Leftrightarrow-\left(2x-5\right).\left(2x+7\right)-\left(2x+5\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(-7-2x-2x-5\right)=0\)
\(\Leftrightarrow\left(2x-5\right)\left(-12-4x\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\-12-4x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)
Vậy x=5/2 hoặc x=-3
S
2
16 tháng 11 2021
\(\left(5x-2\right)\left(2x+7\right)-4x^2-25=0\)
\(10x+35-4x^2-14x-4x^2+25=0\)
\(-4x+60-8x^2=0\)
\(-4\left(2x^2+x-15\right)=0\)
\(-4\left(2x^2+6x-5x-15\right)=0\)
\(-4\left(2x-5\right)\left(x+3\right)=0\)
=> \(x\) ∈ \(\left\{\dfrac{5}{2};-3\right\}\)
T
23 tháng 9 2023
Bài 1.
\(a, (3x-4)^2\)
\(=\left(3x\right)^2-2\cdot3x\cdot4+4^2\)
\(=9x^2-24x+16\)
\(b,\left(1+4x\right)^2\)
\(=1^2+2\cdot1\cdot4x+\left(4x\right)^2\)
\(=16x^2+8x+1\)
\(c,\left(2x+3\right)^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)
\(=8x^3+36x^2+54x+27\)
\(d,\left(5-2x\right)^3\)
\(=5^3-3\cdot5^2\cdot2x+3\cdot5\cdot\left(2x\right)^2-\left(2x\right)^3\)
\(=125-150x+60x^2-8x^3\)
\(e,49x^2-25\)
\(=\left(7x\right)^2-5^2\)
\(=\left(7x-5\right)\left(7x+5\right)\)
\(f,\dfrac{1}{25}-81y^2\)
\(=\left(\dfrac{1}{5}\right)^2-\left(9y\right)^2\)
\(=\left(\dfrac{1}{5}-9y\right)\left(\dfrac{1}{5}+9y\right)\)
Bài 2.
\(a,\left(x-5\right)^2-\left(x+7\right)\left(x-7\right)=8\)
\(\Rightarrow x^2-2\cdot x\cdot5+5^2-\left(x^2-7^2\right)=8\)
\(\Rightarrow x^2-10x+25-\left(x^2-49\right)=8\)
\(\Rightarrow x^2-10x+25-x^2+49=8\)
\(\Rightarrow\left(x^2-x^2\right)-10x=8-25-49\)
\(\Rightarrow-10x=-66\)
\(\Rightarrow x=\dfrac{33}{5}\)
\(b,\left(2x+5\right)^2-4\left(x+1\right)\left(x-1\right)=10\)
\(\Rightarrow\left(2x\right)^2+2\cdot2x\cdot5+5^2-4\left(x^2-1^2\right)=10\)
\(\Rightarrow4x^2+20x+25-4x^2+4=10\)
\(\Rightarrow\left(4x^2-4x^2\right)+20x=10-25-4\)
\(\Rightarrow20x=-19\)
\(\Rightarrow x=\dfrac{-19}{20}\)
#\(Toru\)
KV
23 tháng 9 2023
Bài 1
a) (3x - 4)²
= (3x)² - 2.3x.4 + 4²
= 9x² - 24x + 16
b) (1 + 4x)²
= 1² + 2.1.4x + (4x)²
= 1 + 8x + 16x²
c) (2x + 3)³
= (2x)³ + 3.(2x)².3 + 3.2x.3² + 3³
= 8x³ + 36x² + 54x + 27
d) (5 - 2x)³
= 5³ - 3.5².2x + 3.5.(2x)² - (2x)³
= 125 - 150x + 60x² - 8x³
e) 49x² - 25
= (7x)² - 5²
= (7x - 5)(7x + 5)
f) 1/25 - 81y²
= (1/5)² - (9y)²
= (1/5 - 9y)(1/5 + 9y)
19 tháng 11 2018
\(\left(5-2x\right)\left(2x-7\right)=4x^2-25\)
\(-\left(2x-5\right)\left(2x-7\right)=\left(2x\right)^2-5^2\)
\(\left(2x-5\right)\left(7-2x\right)=\left(2x-5\right)\left(2x+5\right)\)
\(\left(2x-5\right)\left(7-2x\right)-\left(2x-5\right)\left(2x+5\right)=0\)
\(\left(2x-5\right)\left(7-2x-2x-5\right)=0\)
\(\left(2x-5\right)\left(2-4x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-5=0\\2-4x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=\frac{1}{2}\end{cases}}\)
28 tháng 10 2018
Ta có:
\(a)\left(2x-5\right)\left(x+2\right)-2x\left(x-1\right)=15\)
\(\Leftrightarrow\left(2x^2-x-10\right)-\left(2x^2-2x\right)=15\Leftrightarrow x-10=15\)
\(\Leftrightarrow x=25\)
\(b)\left(5-2x\right)\left(2x+7\right)=4x^2-25\)
\(\Leftrightarrow\left(5-2x\right)\left(2x+7\right)=\left(2x-5\right)\left(2x+5\right)\)
\(\Leftrightarrow\left(5-2x\right)\left(4x+12\right)=0\)
\(5-2x=0\Leftrightarrow x=\frac{5}{2}\)
\(4x+12=0\Leftrightarrow x=-3\)
Vậy ..........................................
\(=\left(2x-5\right)\left(2x+5\right)-\left(x+7\right)\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x+5-x-7\right)\)
\(=\left(2x-5\right)\left(x-2\right)\)