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10 tháng 4 2020

a, \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)

\(\Leftrightarrow x^2-4x+4-\left(x^2+6x+9\right)-4x-4=5\)

\(\Leftrightarrow x^2-4x+4-x^2-6x-9-4x-4=5\)

\(\Leftrightarrow-14x-9=5\)

\(\Leftrightarrow-14x=14\)

\(\Leftrightarrow x=-1\)

Vậy....

b, \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(\Leftrightarrow\left(2x\right)^2-3^2-\left(x^2-2x+1\right)-3x^2+15x=-44\)

\(\Leftrightarrow4x^2-9-x^2+2x-1-3x^2+15x=-44\)

\(\Leftrightarrow-10+17x=-44\)

\(\Leftrightarrow17x=-34\)

\(\Leftrightarrow x=-2\)

Vậy....

c, \(\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)

\(\Leftrightarrow\left(5x\right)^2+10x+1-\left[\left(5x\right)^2-3^2\right]=30\)

\(\Leftrightarrow\left(5x\right)^2+10x+1-\left(5x\right)^2+9=30\)

\(\Leftrightarrow10x+10=30\)

\(\Leftrightarrow10x=20\)

\(\Leftrightarrow x=2\)

Vậy....

d, \(\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-2\right)^2=7\)

\(\Leftrightarrow x^2+6x+9+x^2-4-2\left(x^2-4x+4\right)=7\)

\(\Leftrightarrow2x^2+6x+5-2x^2+8x-8=7\)

\(\Leftrightarrow14x-3=7\)

\(\Leftrightarrow14x=10\)

\(\Leftrightarrow x=\frac{10}{14}=\frac{5}{7}\)

Vậy...

12 tháng 4 2020

câu hỏi đây

S
25 tháng 8

\(a.\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)

\(\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-4x-4=5\)

\(\left(-4x-6x\right)+\left(4-9\right)-4x-4=5\)

\(-10x-5-4x-4=5\)

\(-14x-9=5\)

\(-14x=14\Rightarrow x=-1\)

\(b.\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)

\(4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)

\(4x^2-9-x^2+2x-1-3x^2+15x=-44\)

\(17x-10=-44\)

\(17x=-34\Rightarrow x=-2\)

\(c.\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(25x^2+10x+1-\left(25x^2-9\right)=30\)

\(10x+10=30\)

\(10x=20\Rightarrow x=2\)

\(d.\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-1\right)^2=7\)

\(\left(x^2+6x+9\right)+\left(x^2-4\right)-2\left(x^2-2x+1\right)=7\)

\(2x^2+6x+5-2x^2+4x-2=7\)

\(10x+3=7\)

\(10x=4\Rightarrow x=\frac{4}{10}=\frac25\)

\(f.\left(3x-8\right)^2=0\)

\(3x-8=0\Rightarrow x=\frac83\)

\(e.6\left(x+1\right)^2-2\left(x+1\right)+2\left(x-1\right)\left(x^2+x+1\right)=0\)

\(6\left(x^2+2x+1\right)-2x-2+2\left(x^3-1\right)=0\)

\(6x^2+12x+6-2x-2+2x^3-2=0\)

\(2x^3+6x^2+10x+2=0\)

\(\Rightarrow x\approx-0,23\)

7 tháng 8 2017

a)

<=> 10x - 35 + 16x - 10 = 5 

<=> 10x + 16x = 5 + 35 + 10

<=> 26x = 50

<=> x = 50/26 = 25/13

24 tháng 8 2019

\(a,-5x\left(x-3\right)\left(2x+4\right)-\left(x+3\right)\left(x-3\right)+\left(5x-2\right)\left(3x+4\right)\)

\(=-5x\left(2x^2-x-12\right)-\left(x^2-9\right)+15x^2+20x-6x-8\)

\(=-10x^3+5x^2+60x-x^2+9+15x^2+20x-6x-8\)

\(=-10x^3+19x^2+74x+1\)

\(b,\left(4x-1\right)x\left(3x+1\right)-5x^2.x\left(x-3\right)-\left(x-4\right)x\left(x-5\right)\)\(-7\left(x^3-2x^2+x-1\right)\)

\(=\left(4x^2-x\right)\left(3x+1\right)-5x^4-15x^3-\left(x^2-4x\right)\left(x-5\right)\)\(-7x^3+14x^2-7x+7\)

\(=12x^3+x^2-x-5x^4-15x^3-x^3+9x^2+20x\)\(-7x^3+14x^2-7x+7\)

\(=-5x^4-11x^3+24x^2+12x+7\)

\(c,\left(5x-7\right)\left(x-9\right)-\left(3-x\right)\left(2-5x\right)-2x\left(x-4\right)\)

\(=5x^2-52x+63-6+17x-5x^2-2x^2+8x\)

\(=-2x^2-27x+57\)

24 tháng 8 2019

\(d,\left(5x-4\right)\left(x+5\right)-\left(x+1\right)\left(x^2-6\right)-5x+19\)

\(=5x^2+21x-20-x^3-x^2+6x+6-5x+19\)

\(=-x^3+4x^2+22x+5\)

\(e,\left(9x^2-5\right)\left(x-3\right)-3x^2\left(3x+9\right)-\left(x-5\right)\left(x+4\right)-9x^3\)

\(=9x^3-27x^2-5x+15-9x^3-27x^2-x^2+x+20-9x^3\)

\(=-9x^3-55x^2+4x+35\)

\(g,\left(x-1\right)^2-\left(x+2\right)^2\)

\(=x^2-2x+1-x^2-4x-4\)

\(=-6x-3\)

3 tháng 5 2017

c. x^2-5x+6=0

<=> x^2-5x=-6

<=> -4x=-6

<=> x=-6/-4

vậy tập nghiệm của pt là s={-6/-4}