Nếu bn ko nhìn rõ thì ib vs mk nhé, mk sẽ gửi bản full cho bn.

Đấy nhé

có nhìn thấy j đâu  ri msk

a )

a: \(\Leftrightarrow\left|x\cdot\dfrac{1}{3}-\dfrac{1}{4}\right|=\dfrac{1}{2}+\dfrac{1}{6}=\dfrac{3}{6}+\dfrac{1}{6}=\dfrac{2}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{1}{3}-\dfrac{1}{4}=\dfrac{2}{3}\\x\cdot\dfrac{1}{3}-\dfrac{1}{4}=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{11}{4}\\x=-\dfrac{5}{4}\end{matrix}\right.\)

b: \(\Leftrightarrow\left|x\cdot\dfrac{4}{7}+\dfrac{3}{4}\right|=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)

\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{4}{7}+\dfrac{3}{4}=\dfrac{1}{6}\\x\cdot\dfrac{4}{7}+\dfrac{3}{4}=-\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-49}{48}\\x=-\dfrac{77}{48}\end{matrix}\right.\)

a) (3x - 2)² - (1 + 5x)²

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

= (-2x - 3)(8x - 1)

b) (3x + 4)(3x - 4) - (5 - x)²

= (3x)² - 4² - (25 - 10x + x²)

= 9x² - 16 - 25 + 10x - x²

= 8x²  + 10x - 41

c) \(\left(\dfrac{1}{2}x+4\right)^2-\left(\dfrac{1}{2}x+3\right)\left(\dfrac{1}{2}x-3\right)\)

\(=\left(\dfrac{1}{2}x\right)^2+2.\dfrac{1}{2}x.4+4^2-\left[\left(\dfrac{1}{2}x\right)^2-3^2\right]\)

\(=\dfrac{1}{4}x^2+4x+16-\dfrac{1}{4}x^2+9\)

\(=4x+25\)

a: =9x^2-12x+4-25x^2-10x-1

=-16x^2-22x+3

b: =9x^2-16-x^2+10x-25

=8x^2+10x-41

c: \(=\dfrac{1}{4}x^2+4x+16-\dfrac{1}{4}x^2+9=4x+25\)

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

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

\(=-3x^2+7x-4\)

Thay x = 2 vào A được:

\(=-3.2^2+7.2-4=-2\)

Vậy: Giá trị của A khi x = 2 là -2

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b/ \(B=126y^3+\left(x-5y\right)\left(x^2+25y^2+5xy\right)\)

\(=126y^3+x^3-125y^3\)

Thay x = -5 và y = -3 vào B được: 

\(126.\left(-3\right)^3+\left(-5\right)^3-125.\left(-3\right)^3=-152\)

Vậy: Giá trị của B tại x = -5 và y = -3 là -152

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c/ \(C=a^3+b^3-\left(a^2-2ab+b^2\right)\left(a-b\right)\)

\(=a^3+b^3-\left(a-b\right)^3\)

\(=a^3+b^3-a^3+3a^2b-3ab^2+b^3\)

\(=2b^3+3a^2b-3ab^2\)

Thay a = -4 và b = 4 vào C được:

\(2.4^3+3.\left(-4\right)^2.4-3.\left(-4\right).4^2=512\)

Vậy: Giá trị của C tại a = -4 vào b = 4 là 512

a:Ta có: \(A=\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)+3\left(x-1\right)\left(x^2+x+1\right)\)

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

\(=-3x^2+7x-4\)

\(=-3\cdot2^2+7\cdot2-4\)

\(=-12-4+14=-2\)

c: Ta có: \(C=a^3+b^3-\left(a-b\right)\left(a^2-2ab+b^2\right)\)

\(=a^3+b^3-a^3+3a^2b-3ab^2+b^3\)

\(=2b^3+3a^2b-3ab^2\)

\(=2\cdot4^3+3\cdot\left(-4\right)^2\cdot4-3\cdot\left(-4\right)\cdot4^2\)

\(=128+192+192=512\)

a: Ta có: \(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=\dfrac{-7}{x+2}\)

\(\Leftrightarrow3-\left(x+2\right)=-7\left(x-1\right)\)

\(\Leftrightarrow3-x-2+7x-7=0\)

\(\Leftrightarrow6x-6=0\)

hay x=1(loại

b: Ta có: \(\dfrac{2}{-x^2+6x-8}-\dfrac{x-1}{x-2}=\dfrac{x+3}{x-4}\)

\(\Leftrightarrow\dfrac{-2}{\left(x-2\right)\left(x-4\right)}-\dfrac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=\dfrac{\left(x+3\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}\)

Suy ra: \(-2-x^2+5x-4=x^2+x-6\)

\(\Leftrightarrow-x^2+5x-6-x^2-x+6=0\)

\(\Leftrightarrow-2x^2+4x=0\)

\(\Leftrightarrow-2x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)

\(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)

\(\Rightarrow\dfrac{3}{\left(x^2-x\right)+\left(2x-2\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)

\(\Rightarrow\dfrac{3}{x\left(x-1\right)+2\left(x-1\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)

\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{1}{x-1}+\dfrac{7}{x+2}=0\)

\(\Rightarrow\dfrac{3}{\left(x+2\right)\left(x-1\right)}-\dfrac{x+2}{\left(x+2\right)\left(x-1\right)}+\dfrac{7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)

\(\Rightarrow\dfrac{3-\left(x+2\right)+7\left(x-1\right)}{\left(x+2\right)\left(x-1\right)}=0\)

\(\Rightarrow3-x-2+7x-7=0\)

\(\Rightarrow6x-6=0\)

\(\Rightarrow x=1\)

1/2+2/3 = 3/3 = 1

x : 2/5 = 5/6

x = 5/6x2/5

x = 1/3

\(\dfrac{1}{2}+\dfrac{2}{3}=\dfrac{3}{3}=1\)

x : \(\dfrac{2}{5}\) = \(\dfrac{5}{6}\)

x = \(\dfrac{5}{6}\)x\(\dfrac{2}{5}\)

x = \(\dfrac{1}{3}\)

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