\(\sqrt{a}+\sqrt{b}=m\Leftrightarrow m-\sqrt{a}=\sqrt{b}\Rightarrow m^2-2m\sqrt{a}+a=b\)

\(\Leftrightarrow\sqrt{a}=\frac{m^2+a-b}{2m}\)là số hữu tỉ. 

Tương tự cũng suy ra \(\sqrt{b}\)là số hữu tỉ. 

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

\(< =>x^2\left(x^2+4\right)-\left(x^2+4\right)=0\)

\(< =>\left(x^2-1\right)\left(x^2+4\right)=0\)

\(< =>\left(x-1\right)\left(x+1\right)=0\)(vì x² + 4 > 0)

\(< =>\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

Trả lời:

Câu 1: 

a, \(\left(5x+1\right)\left(2x-3\right)-\left(7x-2\right)\left(x+3\right)+5\)

\(=10x^2-15x+2x-3-\left(7x^2+21x-2x-6\right)+5\)

\(=10x^2-13x-3-7x^2-21x+2x+6+5\)

\(=3x^2-32x+8\)

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

\(=9x^2-12x+4+4x^2+20x+25-9x^2+16\)

\(=4x^2+8x+45\)

c, \(\left(3x+2\right)\left(9x^2-6x+4\right)+\left(5-3x\right)\left(25+15x+9x^2\right)-9\)

\(=27x^3+8+125-27x^3-9\)

\(=124\)

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

\(=x^3-9x^2+27x-27-\left(x^2-4\right)\left(x-1\right)\)

\(=x^3-9x^2+27x-27-\left(x^3-x^2-4x+4\right)\)

\(=x^3-9x^2+27x-27-x^3+x^2+4x-4\)

\(=-8x^2+31x-31\)

e, \(\left(x+2y\right)^3-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=x^3+6x^2y+12xy^2+8y^3-\left(8x^3-y^3\right)\)

\(=x^3+6x^2y+12xy^2+8y^3-8x^3+y^3\)

\(=-7x^3+9y^3+6x^2y+12xy^2\)

f, \(\left(x-3y\right)^3-\left(2x-y\right)\left(3x+2y\right)\left(3x-2y\right)\)

\(=x^3-9x^2y+27xy^2-\left(2x-y\right)\left(9x^2-4y^2\right)\)

\(=x^3-9x^2y+27xy^2-27y^3-\left(18x^3-8xy^2-9x^2y+4y^3\right)\)

\(=x^3-9x^2y+27xy^2-27y^3-18x^3+8xy^2+9x^2y-4y^3\)

\(=-17x^3-31y^3+35xy^2\)

Câu 2 : a ) \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Leftrightarrow2x^2-10x-3x-2x^2=-13x=26\Leftrightarrow x=-2\)

b) \(\left(3x+1\right)^2+\left(2x-5\right)^2=13\left(x+2\right)\left(x-2\right)\)

\(\Leftrightarrow9x^2+6x+1+4x^2-20x+25=13\left(x^2-4\right)\)

\(\Leftrightarrow13x^2-14x+26=13x^2-52\Leftrightarrow13x^2-14x+26-13x^2+52=0\)

\(\Leftrightarrow78-14x=0\Leftrightarrow78=14x\Leftrightarrow x=\frac{78}{14}=\frac{39}{7}\)

Sửa đề: (x-3)(x² + 3x+9) - x(x-4)(x+4)=2

<=> (x³ - 3³) - x(x² - 4²) =2

<=> x³ -27 - x³ + 16x =2

<=> 16x = 2 + 27

<=> 16x = 29

<=> x = 29/16

Vậy x = 29/16

( x - 3 )( x² + 3x + 9 ) + x( x - 4 )( x + 4 ) = 2

<=> x³ - 27 + x( x² - 4 ) - 2 = 0

<=> x³ - 27 + x³ - 4x - 2 = 0

<=> 2x³ - 4x - 29 = 0

đến đây chịu:)

Trả lời:

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

\(=x^3-6x^2+12x-8-\left(x^3+1\right)+6\left(x^2-2x+1\right)\)

\(=x^3-6x^2+12x-8-x^3-1+6x^2-12x+6\)

\(=-2\)

2, \(-x\left(x+2\right)^2+\left(2x+1\right)^2+\left(x+3\right)\left(x^2-3x+9\right)-1\)

\(=\)\(-x\left(x^2+4x+4\right)+4x^2+4x+1+x^3+27-1\)

\(=-x^3-4x^2-4x+4x^2+4x+1+x^3+27-1\)

\(=27\)

xin lỗi nha, ý thứ nhất kết quả là - 3, mình đánh máy nhầm

Trả lời:

\(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)

\(=\left(2bc\right)^2-\left(b^2+c^2-a^2\right)^2\)

\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)

\(=\left[a^2-\left(b^2-2bc+c^2\right)\right]\left[\left(b^2+2bc+c^2\right)-a^2\right]\)

\(=\left[a^2-\left(b-c\right)^2\right]\left[\left(b+c\right)^2-a^2\right]\)

\(=\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(b+c+a\right)\)

\(4b^2c^2-\left(b^2+c^2-a^2\right)^2\)

\(=\left(2bc-b^2-c^2+a^2\right)\left(2bc+b^2+c^2-a^2\right)\)

\(=\left[a^2-\left(b-c\right)^2\right]\left[\left(b+c\right)^2-a^2\right]\)

\(=\left(a-b+c\right)\left(a+b-c\right)\left(b+c-a\right)\left(a+b+c\right)\)