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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
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
a. \(9\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow9x+18-3x-6=0\)
\(\Leftrightarrow6x+12=0\)
\(\Leftrightarrow x=-2\)
e. \(\left(2x-1\right)^2-45=0\)
\(\Leftrightarrow4x^2-2x+1-45=0\)
\(\Leftrightarrow4x^2-2x-44=0\)
Đến đó tự giải tiếp nha!
c. \(2\left(2x-5\right)-3x=0\)
\(\Leftrightarrow4x-10-3x=0\)
\(\Leftrightarrow x-10=0\)
\(\Leftrightarrow x=10\)
g. \(2x^2-6x=0\)
\(\Leftrightarrow2x\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
a/ \(\left(2x-3\right)^2-\left(3x+2\right)^2=5x\left(2-x\right)\)
<=> \(\left(2x-3-3x-2\right)\left(2x-3+3x+2\right)=5x\left(2-x\right)\)
<=> \(\left(-x-5\right)\left(5x-1\right)=5x\left(2-x\right)\)
<=> \(-5x^2-25x+x+5=10x-5x^2\)
<=> \(10x+25x-x=5\)
<=> \(34x=5\)
<=> \(x=\frac{5}{34}\)
b/ pt <=> \(2^3x^3-3.2^2.x^2.1+3.2.x.1^2-1^3=0\)
<=> \(\left(2x-1\right)^3=0\)
<=> 2 x - 1 = 0
<=> x = 1/2.
1) 3(x - 1)2 - 3x(x - 5) = 1
⇒ 3(x2 - 2x + 1) - 3x2 + 15x = 1
⇒ 3x2 - 6x + 3 - 3x2 + 15x = 1
⇒ 9x = 1 - 3
⇒ 9x = -2
⇒ x = \(\dfrac{-2}{9}\)
2) (6x−2)2+(5x−2)2−4(3x−1)(5x−2)=0
⇒ (6x - 2)2 + (5x - 2)2 -4(6x - 2)(5x - 2) = 0
⇒ (6x - 2)2 -2(6x - 2)(5x - 2) + (5x - 2)2 -2(6x - 2)(5x - 2) = 0
⇒ (6x - 2)(6x - 2 - 5x +2) + (5x - 2)(5x - 2 - 6x + 2) = 0
⇒ x(6x - 2) - x(5x - 2) = 0
⇒ x(6x - 2 - 5x +2) = 0
⇒ xx = 0
⇒ x = 0
Còn mấy cái sau mình trả lời sau nha 
Còn hai câu sau nữa nè :)
3) (2x - 5)(2x + 5) - 1 = 0
⇒ 4x2 - 25 - 1 = 0
⇒ 4x2 = 26
⇒ x2 = \(\dfrac{13}{2}\)
⇒ x = \(\sqrt{\dfrac{13}{2}}\) hoặc x = -\(\sqrt{\dfrac{13}{2}}\)
4) 5x2 - 20 = 0
⇒ 5x2 = 20
⇒ x2 = 4
⇒ x = 2 hoặc x = -2
\(\left(x+1\right)^2=4\left(x^2-2x+1\right)^2\\\Leftrightarrow\left(x+1\right)^2=4\left(x-1\right)^2\\\Leftrightarrow \left(x+1\right)^2-4\left(x-1\right)^2=0\\\Leftrightarrow \left(x+1\right)^2-\left(2x-2\right)^2=0\\\Leftrightarrow \left[\left(x+1\right)+\left(2x-2\right)\right]\left[\left(x+1\right)-\left(2x-2\right)\right] =0\\ \Leftrightarrow\left(x+1+2x-2\right)\left(x+1-2x+2\right)=0\\\Leftrightarrow \left(3x-1\right)\left(3-x\right)=0\\\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=3\end{matrix}\right. \)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{1}{3};3\right\}\)
\(\left(2x+7\right)^2=9\left(x+2\right)^2\\ \Leftrightarrow\left(2x+7\right)^2-9\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+7\right)^2-\left(3x+6\right)^2=0\\ \Leftrightarrow\left[\left(2x+7\right)+\left(3x+6\right)\right]\left[\left(2x+7\right)-\left(3x+6\right)\right]=0\\ \Leftrightarrow\left(2x+7+3x+6\right)\left(2x+7-3x-6\right)=0\\ \Leftrightarrow\left(5x+13\right)\left(1-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x+13=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-13}{5}\\x=1\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-13}{5};1\right\}\)
\(4\left(2x+7\right)^2=9\left(x+3\right)^2\\\Leftrightarrow 4\left(2x+7\right)^2-9\left(x+3\right)=0\\ \Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\\\Leftrightarrow \left[\left(4x+14\right)+\left(3x+9\right)\right]\left[\left(4x+14\right)-\left(3x+9\right)\right]=0\\\Leftrightarrow \left(4x+14+3x+9\right)\left(4x+14-3x-9\right)=0\\\Leftrightarrow \left(7x+23\right)\left(x+5\right)=0\\\Leftrightarrow\left[{}\begin{matrix}7x+23=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-23}{7}\\x=-5\end{matrix}\right. \)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-23}{7};-5\right\}\)
a, (3x - 5)(2x - 1) - (x + 2)(6x - 1) = 0
=> 6x^2 - 3x - 10x + 5 - (6x^2 - x + 12x - 2) = 0
=> 6x^2 - 13x + 5 - 6x^2 - 11x + 2 = 0
=> -24x + 7 = 0
=> - 24x = -7
=> x = 7/24
b, (3x - 2)(3x + 2) - (3x - 1)^2 = -5
=> 9x^2 - 4 - 9x^2 + 6x - 1 = -5
=> 6x - 5 = -5
=> 6x = 0
=> x = 0
c, x^2 = -6x - 8
=> x^2 + 6x + 8 = 0
=> x^2 + 2.x.3 + 9 - 1 = 0
=> (x + 3)^2 = 1
=> x + 3 = 1 hoặc x + 3 = -1
=> x = -2 hoặc x = -4
a/ Sai đề à??
\(\left(2x^3-3\right)^2-\left(4x^2-9\right)=0\)
\(\Leftrightarrow4x^6-12x^3+9-4x^2+9=0\)
\(\Leftrightarrow4x^6-13x^2-4x^2+18=0\)
b/ \(\Leftrightarrow\left(x^2-3\right)\left(x^2+3\right)+2x\left(x^2-3\right)=0\)
\(\Leftrightarrow\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(x^2+3+2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\) (do \(x^2+3+2x>0\forall x\))
d/ \(\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
x2 - 5x - 36 = 0
=> x2 - 9x + 4x - 36 = 0
=> x(x - 9) + 4(x - 7) = 0
=> (x + 4)(x - 7) = 0
=> \(\orbr{\begin{cases}x+4=0\\x-7=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-4\\x=7\end{cases}}\)
6x2 - (2x + 5)(3x - 2) = -12
=> 6x2 - 6x2 + 4x - 15x + 10 = -12
=> -11x = -22
=> x = 2
x2 - 25 = 6x - 9
=> x2 - 25 - 6x + 9 = 0
=> x2 - 6x - 16 = 0
=> x2 - 8x + 2x - 16 = 0
=> x(x - 8) + 2(x - 8) = 0
=> (x + 2)(x - 8) = 0
=> \(\orbr{\begin{cases}x+2=0\\x-8=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-2\\x=8\end{cases}}\)