a) B= 2x²-3x+1

=(2x²-2x)-(x-1)

=2x(x-1)-(x-1)

=(2x-1)(x-1)

\(\left|x\right|=\frac{1}{2}\)nên ta có \(x=\frac{1}{2}\)hoặc\(x=\frac{-1}{2}\)

nếu \(x=\frac{1}{2}\)thì

B=(2*\(\frac{1}{2}\)-1)(\(\frac{1}{2}\)-1)

B=0

nếu x= -1/2

thì B= (2*(-1/2)-1)(-1/2-1)

B=(-2)*(-3/2)

B=3

giúp e vs các a cj soyeon_Tiểubàng giải

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Võ Đông Anh Tuấn

a) \(\left(3x-5\right)\left(2x+3\right)-\left(2x-3\right)\left(3x+7\right)-2x\left(x-4\right)\)

\(=\left(6x^2-x-15\right)-\left(6x^2+5x-21\right)-\left(2x^2-8x\right)\)

\(=6x^2-x-15-6x^2-5x+21-2x^2+8x\)

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

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

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

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

\(=\left(x^2+2\right)^2-\left(x^4-16\right)\)

\(=\left(x^4+4x^2+4\right)-\left(x^4-16\right)\)

\(=x^4+4x^2+4-x^4+16\)

\(=4x^2+20\)

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

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

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

\(=4x^2-4xy+y^2-2x^2-16xy-18y^2-9x^2+1\)

\(=-7x^2-20xy-17y^2+1\)

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

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

\(=x^6-3x^4+3x^2-1-x^6+1\)

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

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

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

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

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

\(=\left[\left(2x-1\right)-\left(2x+1\right)\right]^2\)

\(=\left(2x-1-2x-1\right)^2\)

\(=\left(-2\right)^2=4\)

g) \(\left(x-y+z\right)^2+\left(y-z\right)^2-2\left(x-y+z\right)\left(z-y\right)\)

\(=\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)

\(=\left(x-y+z+y+z\right)^2\)

\(=\left(x+2z\right)^2\)

h) \(\left(2x+3\right)^2+\left(2x+5\right)^2-\left(4x+6\right)\left(2x+5\right)\)

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

\(=\left[\left(2x+3\right)-\left(2x+5\right)\right]^2\)

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

\(=\left(-2\right)^2=4\)

i) \(5x^2-\dfrac{10x^3+15x^2-5x}{-5x}-3\left(x+1\right)\)

\(=5x^2-\dfrac{-5x\left(-2x^2-3x+1\right)}{-5x}-3\left(x+1\right)\)

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

\(=5x^2+2x^2+3x-1-3x-3\)

\(=7x^2-4\)

a/ +) \(\dfrac{x}{3}=\dfrac{y}{4}\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12}\)\(\left(1\right)\)

+) \(\dfrac{y}{3}=\dfrac{z}{5}\Leftrightarrow\dfrac{y}{12}=\dfrac{z}{20}\left(2\right)\)

Từ \(\left(1\right)+\left(2\right)\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\)

\(\Leftrightarrow\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}\)

Theo t/c dãy tỉ số bằng nhau ta có :

\(\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}=\dfrac{2x-3y+z}{18-36+20}=\dfrac{6}{2}=3\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{9}=3\\\dfrac{y}{12}=3\\\dfrac{z}{20}=3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=27\\y=36\\z=60\end{matrix}\right.\)

Vậy ..

b/ \(2x=3y=5z\)

\(\Leftrightarrow\dfrac{2x}{30}=\dfrac{3y}{30}=\dfrac{5z}{30}\)

\(\Leftrightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)

Theo t/c dãy tỉ số bằng nhau tcos :

\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x+y-z}{15+10-6}=\dfrac{95}{19}=5\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{15}=5\\\dfrac{y}{10}=5\\\dfrac{z}{6}=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=75\\y=50\\z=30\end{matrix}\right.\)

Vậy..

c/ tương tự

1,Thực hiện phép tính :

a, (x + 2)⁹ : (x + 2)⁶

=(x+2)^9-6

=(x+2)³

b, (x - y) ⁴ : (x - 2)³

=(x-y)^4-3

=x-y

c, ( x²+ 2x + 4)⁵ : (x² + 2x + 4)

=(x²+2x+4)^5-1

=(x²+2x+4)⁴

d, 2(x² + 1)³ : 1/3(x² + 1)

=(2÷1/3).[(x²+1)³÷(x²+1)]

=6(x²+1)²

e, 5 (x - y)⁵ : 5/6 (x - y)²

=(5÷5/6).[(x-y)⁵÷(x-y)²]

=6(x-y)⁾³

\(3+5{y}\)

b, x² +y²+z² +2x-4y-6z+14=0

<=> (x²+2x+1)+(y²-4y+4)+(z²-6z+9)=0

<=> (x+1)²+(y-2)²+(z-3)²=0

=>(x+1)²=(y-2)²=(z-3)²=0

=>x+1=y-2=z-3=0

=> x=-1; y=2; z=3

c, 2x²+y²-6x-4y+2xy+5=0

<=> (x²+y²+4+2xy-4x-4y)+(x²-2x+1)=0

<=> (x+y-2)²+(x-1)²=0

=> (x+y-2)²=(x-1)²=0

=>x+y-2=x-1=0

=>x=1; y=1

a/ \(2^{x+1}.3y=12^x\Leftrightarrow2^x.2.3y=2^x.2^x.3^x\Leftrightarrow2^{x-1}.3^{x-1}=y\Leftrightarrow y=6^{x-1}\)

Vậy \(\left(x;y\right)=\left(x;6^{x-1}\right)\)

b/ \(10^x:5^y=20^y\Leftrightarrow10^x=20^y.5^y\Leftrightarrow10^x=100^y\Leftrightarrow10^x=10^{2y}\Leftrightarrow x=2y\)

Vậy \(\left(x;y\right)=\left(2y;y\right)\)

c/ Ta có : \(2^x=4^{y-1}\Leftrightarrow2^x=2^{2y-2}\Leftrightarrow x=2y-2\)

Lại có : \(27y=3^{x+8}\Leftrightarrow27y=3^{x+5}.27\Leftrightarrow y=3^{x+5}\)

Suy ra hệ : \(\begin{cases}x=2y-2\\y=3^{x+5}\end{cases}\)