Question

[LIÊN]

B1 ( kết quả thôi ko cần lời giải)

a) \(\left(4x-3\right)\left(3x+2\right)-\left(6x+1\right)\left(2x-5\right)+1\)

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

c) \(\left(2x+1\right)\left(4x^2-2x+1\right)+\left(2-3x\right)\left(4+6x+9x^2\right)-9\)

B2 tìm x(kết quả)

a) \(3x\left(x-4\right)-x\left(5+3x\right)=-34\)

b) \(\left(3x+1\right)^2+\left(5x-2\right)^2=34\left(x+2\right)\left(x-2\right)\)

c) \(x^3+3x^2+3x+28=0\)

Answer 1

[Lightning Farron]

Bài 1:

a)(4x-3)(3x+2)-(6x+1)(2x-5)+1

=12x²-x-6-12x²+28x+5+1

=27x

b)(3x+4)²+(4x-1)²+(2+5x)(2-5x)

=9x²+24x+16+16x²-8x+1+4-25x²

=16x+21

c)(2x+1)(4x²-2x+1)+(2-3x)(4+6x+9x²)-9

=8x³+1+8-27x³-9

=-19x³

Answer 2

[Lightning Farron]

Bài 2:

a)3x(x-4)-x(5+3x)=-34

=>3x²-12x-3x²-5x=-34

=>-17x=-34

=>x=2

Vậy x=2

b)(3x+1)²+(5x-2)²=34(x+2)(x-2)

=>9x²+6x+1+25x²-20x+4=34(x²-4)

=>34x²-14x+5-34x²+136=0

=>-14x+141=0

=>-14x=-141

=>x=\(\frac{141}{14}\)

Vậy x=\(\frac{141}{14}\)

c)x³+3x²+3x+28=0

=>x³-x²+7x+4x²-4x+28=0

=>x(x²-x+7)+4(x²-x+7)=0

=>(x+4)(x²-x+7)=0

\(\Rightarrow\left[\begin{array}{nghiempt}x+4=0\\x^2-x+7=0\left(2\right)\end{array}\right.\)

\(\Rightarrow\left[\begin{array}{nghiempt}x=-4\\\left(2\right)\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{27}{4}>0\end{array}\right.\)

=>(2) vô nghiệm

Vậy x=-4