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\(A=\dfrac{3}{2\left(2x-1\right)}\cdot x^2\left|2x-1\right|\cdot2\sqrt{2}\)
\(=\pm3\sqrt{2}x^2\)
\(B=\dfrac{a-b}{b^2}\cdot\dfrac{b^2\cdot\left|a\right|}{\left|a-b\right|}\)
\(=\pm\left|a\right|\)
Câu a, bạn coi lại đề xem $a^2=6-3\sqrt{3}$ hay $a=6-3\sqrt{3}$???
b.
\(B=\frac{\sqrt{(x-2)+(x+2)+2\sqrt{(x-2)(x+2)}}}{\sqrt{x^2-4}+x+2}\)
\(=\frac{\sqrt{(\sqrt{x-2}+\sqrt{x+2})^2}}{\sqrt{x^2-4}+x+2}=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x^2-4}+x+2}=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x+2}(\sqrt{x-2}+\sqrt{x+2})}=\frac{1}{\sqrt{x+2}}\)
\(=\frac{1}{\sqrt{3+\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{6+2\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{(\sqrt{5}+1)^2}}=\frac{\sqrt{2}}{\sqrt{5}+1}\)
\(b.\)
\(=\sqrt{\left(3a\right)^2\cdot\left(b-2\right)^2}\)
\(=\left|3a\right|\cdot\left|b-2\right|\)
Với : \(a=2,b=-\sqrt{3}\)
\(2\cdot3\cdot\left(-\sqrt{3}-2\right)=6\cdot\left(-\sqrt{3}-2\right)\)
a: =xy(1/3+4-2)=7/3xy
b: =xy^2(-1+3/2+4/3)=(1/3+3/2)xy^2=11/6xy^2
c: =4x^2y^2+2/3x^2y^2-4/3x^2y=-4/3x^2y+14/3x^2y^2
d: =3x^2y^2z+4x^2y^2z-8x^2y^2z=-x^2y^2z
a) \(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}=\left|3a^2\right|=3a^2\)
b) \(2\sqrt{a^2}-5a=2\left|a\right|-5a=-2a-5a=-7a\)
c) \(\sqrt{16\left(1+4x+4x^2\right)}=\sqrt{\left[4\left(1+2x\right)\right]^2}=\left|4\left(1+2x\right)\right|=4\left(1+2x\right)\)
Bài 1:
a: Ta có: \(A=\left(k-4\right)\left(k^2+4k+16\right)-\left(k^3+128\right)\)
\(=k^3-64-k^3-128\)
=-192
b: Ta có: \(B=\left(2m+3n\right)\left(4m^2-6mn+9n^2\right)-\left(3m-2n\right)\left(9m^2+6mn+4n^2\right)\)
\(=8m^3+27n^3-27m^3+8n^3\)
\(=-19m^3+35n^3\)
Bài 4:
a: Ta có: \(\left(x-1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=16\)
\(\Leftrightarrow x^3-3x^2+3x-1+8-x^3+3x^2+6x=16\)
\(\Leftrightarrow9x=9\)
hay x=1
b: ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)
\(\Leftrightarrow x^3+8-x^3+2x=15\)
\(\Leftrightarrow2x=7\)
hay \(x=\dfrac{7}{2}\)
a) Ta có A = 8 ( a 2 + b 2 ) a ( a 2 − 16 b 2 ) . a 2 − 16 b 2 a 2 + b 2 = 8 a
b) Ta có B = 2 t + 2 t + 2 . 4 − t 2 4 − 4 t 2 = 2 − t 2 − 2 t