a) \(\left(2x-5\right)^2-\left(2x+3\right)\left(2x-3\right)=10\Leftrightarrow\left(4x^2-20x+25\right)-\left(4x^2-9\right)-10=0\)

\(\Leftrightarrow-20x+24=0\Leftrightarrow x=\frac{6}{5}\)

b) \(\left(4x-1\right)\left(x+2\right)-\left(2x+3\right)^2-5\left(x-1\right)=9\Leftrightarrow-10x-15=0\)

\(\Leftrightarrow x=\frac{-3}{2}\)

c) \(\left(x+1\right)^3-\left(x-1\right)^3-2=6\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-8=0\)

\(\Leftrightarrow6x^2-6=0\Leftrightarrow x=\pm1\)

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

\(\Leftrightarrow\left(x^3+8\right)-\left(x^3+1\right)+3x+6=5\Leftrightarrow3x+8=0\Leftrightarrow x=\frac{-8}{3}\)

Câu a :

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

\(=4x^2+4x+1-4x^2+20\)

\(=4x+19\)

Câu b :

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

\(=x^2+6x+9-x^2-1\)

\(=6x-8\)

Câu c :

\(\left(x-5\right)^2-\left(x+2\right)^2\)

\(=\left(x-5-x-2\right)\left(x-5+x+2\right)\)

\(=-7\left(2x-3\right)\)

Bạn trình bày cho rõ ràng xem nào.

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

= 2( x² - 2x + 1 ) - 4( 9 + 6x + x² ) + 2x² - 10x

= 2x² - 4x + 2 - 36 - 24x - 4x² + 2x² - 10x

= ( 2x² - 4x² + 2x² ) + ( -4x - 24x - 10x ) + ( 2 - 36 )

= -38x - 34

b) 2( 2x + 5 )² - 3( 4x + 1 )( 1 - 4x )

= 2( 4x² + 20x + 25 ) + 3( 4x + 1 )( 4x - 1 )

= 8x² + 40x + 50 + 3( 16x² - 1 )

= 8x² + 40x + 50 + 48x² - 3

= 56x² + 40x + 47

c) ( x - 1 )³ - x( x - 3 )² + 1

= x³ - 3x² + 3x - 1 - x( x² - 6x + 9 ) + 1

= x³ - 3x² + 3x - x³ + 6x² - 9x

= 3x² - 6x

d) ( x + 2 )³ - x²( x + 6 ) 

= x³ + 6x² + 12x + 8 - x³ - 6x²

= 12x + 8

e) ( x - 2 )( x + 2 ) - ( x + 1 )³ - 2x( x - 1 )²

= x² - 4 - ( x³ + 3x² + 3x + 1 ) - 2x( x² - 2x + 1 )

= x² - 4 - x³ - 3x² - 3x - 1 - 2x³ + 4x² - 2x

= -3x³ + 2x² - 5x - 5 

f) ( a + b - c )² - ( b - c )² - 2a( b - c )

= [ ( a + b ) - c ]² - ( b² - 2bc + c² ) - 2ab + 2ac

= [ ( a + b )² - 2( a + b )c + c² ] - b² + 2bc - c² - 2ab + 2ac

= a² + 2ab + b² - 2ac - 2bc + c² - b² + 2bc - c² - 2ab + 2ac

= a²

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

Dùng hẳng đẳng thức thứ nhất + hai :

= \(2\left(x^2-2\cdot x\cdot1+1^2\right)-4\left(3^2+2\cdot3\cdot x+x^2\right)+2x^2-10x\)

= \(2\left(x^2-2x+1\right)-4\left(9+6x+x^2\right)+2x^2-10x\)

= \(2x^2-4x+2-36-24x-4x^2+2x^2-10x\)

= \(-38x-34\)

b) 2(2x + 5)² - 3(4x + 1)(1 - 4x)

Dùng đẳng thức thứ 1 + 3

= 2[(2x)² + 2.2x.5 + 5² ] - (-3)[(4x)² - 1² ]

= 2(4x² + 20x + 25) - (-3).(16x² - 1)

= 8x² + 40x + 50 - (3 - 48x²)

= 8x² + 40x + 50 - 3 + 48x²

= 56x² + 40x + 47

c) (x - 1)³ - x(x - 3)² + 1

Dùng đẳng thức 2 + 5:

= x³ - 3.x².1 + 3.x.1² - 1³ - x(x² - 2.x.3 + 3²) + 1

= x³ - 3x² + 3x - 1 - x³ + 6x² - 9x + 1

= (x³ - x³) + (-3x² + 6x²) + (3x - 9x) + (-1 + 1)

= 3x² - 6x

d) (x + 2)³ - x²(x + 6)

= x³ + 3.x².2 + 3.x.2² + 2³ - x³ - 6x²

= x³ + 6x² + 12x + 8 - x³ - 6x²

= (x³ - x³) + (6x² - 6x²) + 12x + 8 = 12x + 8

e) Dùng đẳng thức thứ 3,4 và 2

= x² - 4 - (x³ + 3.x².1 + 3.x.1² + 1³) - 2x(x² - 2.x.1 + 1²)

= x² - 4 - (x³ + 3x² + 3x + 1) - 2x³ + 4x² - 2x

= x² - 4 - x³ - 3x² - 3x - 1 - 2x³ + 4x² - 2x

= (x² - 3x² + 4x²) + (-4 - 1) + (-x³ - 2x³) + (-3x - 2x)

= 2x² - 5 - 3x³ - 5x

f) Đặt \(a+b-c=A\)

\(b-c=B\)

= \(A^2-B^2-2AB\)

= \(A^2-2AB+\left(-B\right)^2\)

\(=A^2-2AB+B^2\)

= (A - B)²

= (a + b - c - (b - c))²

= (a + b - c - b + c)²

= a²

a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)

\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)

\(\Leftrightarrow2x=-40\)

\(\Rightarrow x=-20\)

b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)

\(\Leftrightarrow x^3+27-x^3+4x=15\)

\(\Leftrightarrow4x=-12\)

\(\Rightarrow x=-3\)

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

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

\(\Leftrightarrow-14x=14\)

\(\Rightarrow x=-1\)

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

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

\(\Leftrightarrow17x=-34\)

\(\Rightarrow x=-2\)

e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Leftrightarrow24x=24\)

\(\Rightarrow x=1\)

Lời giải:
a. $x^3=4^3\Rightarrow x=4$

b. $x^2=49=7^2=(-7)^2$

$\Rightarrow x=7$ hoặc $x=-7$

c. $x^3+1=28$

$x^3=28-1=27=3^3$

$\Rightarrow x=3$

d. $2^x=16=2^4$

$\Rightarrow x=4$

e. $2^4.2^x=2^6$

$\Rightarrow 2^{4+x}=2^6$

$\Rightarrow 4+x=6$

$\Rightarrow x=2$

g.

$5^x=25.5^3=5^2.5^3=5^5$

$\Rightarrow x=5$

Lần sau bạn lưu ý viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để đề được rõ ràng hơn nhé.

Bài 4:

a: |x-2|=0

=>x-2=0

=>x=2

b: |x+3|=1

=>\(\left[\begin{array}{l}x+3=1\\ x+3=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=1-3=-2\\ x=-1-3=-4\end{array}\right.\)

c: \(-3\left|4-x\right|=-9\)

=>|4-x|=3

=>|x-4|=3

=>\(\left[\begin{array}{l}x-4=3\\ x-4=-3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=7\\ x=1\end{array}\right.\)

d: |2x+1|=-2

mà -2<0<=|2x+1|

nên x∈∅

Bài 5:

a: \(\left(x+3\right)^2=36\)

=>\(\left[\begin{array}{l}x+3=6\\ x+3=-6\end{array}\right.\Rightarrow\left[\begin{array}{l}x=6-3=3\\ x=-6-3=-9\end{array}\right.\)

b: \(\left(x+5\right)^2=100\)

=>\(\left[\begin{array}{l}x+5=10\\ x+5=-10\end{array}\right.\Rightarrow\left[\begin{array}{l}x=10-5=5\\ x=-10-5=-15\end{array}\right.\)

c: \(\left(2x-4\right)^2=0\)

=>2x-4=0

=>2x=4

=>x=2

d: \(\left(x-1\right)^3=27\)

=>x-1=3

=>x=4

Bài 8:

a: \(\left(\dfrac{2}{5}+\dfrac{3}{4}\right)^2=\left(\dfrac{8+15}{20}\right)^2=\left(\dfrac{23}{20}\right)^2=\dfrac{529}{400}\)

b: \(\left(\dfrac{5}{4}-\dfrac{1}{6}\right)^2=\left(\dfrac{15}{12}-\dfrac{2}{12}\right)^2=\left(\dfrac{13}{12}\right)^2=\dfrac{169}{144}\)

=0 bạn nha

a,(\(6x-5x^2-15+2x^3:\left(2x-5\right)\)

\(\left(2x^3-5x^2+6x-15\right):\left(2x-5\right)\)