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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
\(2x^2-4x=2x\left(x-2\right)\)
\(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)
\(10\left(x-y\right)-6x\left(y-x\right)=10\left(x-y\right)+6x\left(x-y\right)=\left(10+6x\right)\left(x-y\right)=2\left(x-y\right)\left(3x+5\right)\)\(\left(x+1\right)^2-25=\left(x+1+5\right)\left(x+1-5\right)=\left(x+6\right)\left(x-4\right)\)
\(x^2+3x-y^2+3y=\left(x-y\right)\left(x+y\right)+3\left(x+y\right)=\left(x+y\right)\left(x-y+3\right)\)
\(3x^2+5y-3xy-5x=3x\left(x-y\right)-5\left(x-y\right)=\left(3x-5\right)\left(x-y\right)\)
\(x^2-7x-y^2+7y=\left(x-y\right)\left(x+y\right)-7\left(x-y\right)=\left(x-y\right)\left(x+y-7\right)\)
\(3y^2-3z^2+3x^2=3\left(y^2-z^2+x^2\right)\)
a) Ta có: \(\left(6x-2\right)^2+\left(5x-2\right)^2-4\left(3x-1\right)\left(5x-2\right)=0\)
\(\Leftrightarrow\left(6x-2\right)^2-2\cdot\left(6x-2\right)\left(5x-2\right)+\left(5x-2\right)^2=0\)
\(\Leftrightarrow\left(6x-2-5x+2\right)^2=0\)
\(\Leftrightarrow x^2=0\)
hay x=0
Vậy: x=0
b) Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x\left(x^2-4\right)-5=0\)
\(\Leftrightarrow x^3-6-x^2+4x=0\)
\(\Leftrightarrow4x-6=0\)
\(\Leftrightarrow4x=6\)
hay \(x=\frac{3}{2}\)
Vậy: \(x=\frac{3}{2}\)
c) Ta có: \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x^2-4\right)=2\)
\(\Leftrightarrow x^3-3x^2+3x-1-\left(x^3+27\right)+3x^2-12-2=0\)
\(\Leftrightarrow x^3+3x-15-x^3-27=0\)
\(\Leftrightarrow3x-42=0\)
\(\Leftrightarrow3x=42\)
hay x=14
Vậy: x=14
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.
\(a,\left(2x^2+1\right)+4x>2x\left(x-2\right)\)
\(\Leftrightarrow2x^2+1+4x>2x^2-4x\)
\(\Leftrightarrow4x+4x>-1\)
\(\Leftrightarrow8x>-1\)
\(\Leftrightarrow x>-\frac{1}{8}\)
\(b,\left(4x+3\right)\left(x-1\right)< 6x^2-x+1\)
\(\Leftrightarrow4x^2-4x+3x-3< 6x^2-x+1\)
\(\Leftrightarrow4x^2-x-3< 6x^2-x+1\)
\(\Leftrightarrow4x^2-6x^2< 1+3\)
\(\Leftrightarrow-2x^2< 4\)
\(\Leftrightarrow x^2>2\)
\(\Leftrightarrow x>\pm\sqrt{2}\)
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
1/ (5x+2)2+(6x-3y)2=0
Ta nhận thấy: (5x+2)2\(\ge\)0 và (6x-3y)2\(\ge\)0
Tổng của 2 số dương bằng 0 khi và chỉ khi cả 2 số đều bằng 0
=> \(\hept{\begin{cases}\left(5x+2\right)^2=0\\\left(6x-3y\right)^2=0\end{cases}}< =>\hept{\begin{cases}5x+2=0\\2x-y=0\end{cases}}\)
=> \(\hept{\begin{cases}x=-\frac{2}{5}\\y=2x=-\frac{4}{5}\end{cases}}\)
2/ Làm tương tự 1:
\(\hept{\begin{cases}\left(x+2\right)^2=0\\\left(3x-7y\right)^2=0\end{cases}}< =>\hept{\begin{cases}x+2=0\\3x-7y=0\end{cases}}\)
=> \(\hept{\begin{cases}x=-2\\y=\frac{3x}{7}=-\frac{6}{7}\end{cases}}\)