1: =>x=8-35=-27

2: =>15-4+x=6

=>x+11=6

hay x=-5

3: =>-30+25-x=-1

=>x+5=1

hay x=-4

4: =>x-(-13)=-8

=>x+13=-8

hay x=-21

5: =>x-29-17+38=-9

=>x-8=-9

hay x=-1

G(x) = 8(x + 1)³ + 1

G(x) = 0

⇒ 8(x + 1)³ + 1 = 0

8(x + 1)³ = -1

(x + 1)³ = -1/8

(x + 1)³ = (-1/2)³

x + 1 = -1/2

x = -1/2 - 1

x = -3/2

Vậy nghiệm của G(x) là x = -3/2

H(x) = 8/9 - 2((x - 1)²

H(x) = 0

⇒ 8/9 - 2(x - 1)² = 0

2(x - 1)² = 8/9

(x - 1)² = 8/9 : 2

(x - 1)² = 4/9

x - 1 = 2/9 hoặc x - 1 = -2/9

*) x - 1 = 2/9

x = 2/9 + 1

x = 11/9

*) x - 1 = -2/9

x = -2/9 + 1

x = 7/9

Vậy nghiệm của H(x) là x = 7/9; x = 11/9

K(x) = (x + 8)(x² - 9/25)

K(x) = 0

⇒ (x + 8)(x² - 9/25) = 0

⇒ x - 8 = 0 hoặc x² - 9/25 = 0

*) x - 8 = 0

x = 8

*) x² - 9/25 = 0

x² = 9/25

x = 3/5 hoặc x = -3/5

Vậy nghiệm của K(x) là x = -3/5; x = 3/5; x = 8

a) \(7x\left(x+1\right)-3\left(x+1\right)=0\Rightarrow\left(x+1\right)\left(7x-3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x+1=0\\7x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{7}\end{matrix}\right.\)

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => \(\left[{}\begin{matrix}x+8=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-8\\x=3\end{matrix}\right.\)

c) \(x^2-10x=-25\Rightarrow x^2-10x+25=0\Rightarrow\left(x-5\right)^2=0\Rightarrow x=5\)

d) Giống câu c

⇒[x+1=07x+3=0⇒⎡⎣x=−1x=−37⇒[x+1=07x+3=0⇒[x=−1x=−37

b) 3(x + 8) - x² - 8x = 0

=> 3(x + 8) - (x² + 8x) = 0

=> 3(x + 8) - x(x + 8) = 0

=> (x + 8)(3 - x) = 0 => [x+8=03−x=0⇒[x=−8x=3[x+8=03−x=0⇒[x=−8x=3

c) ${}^{}-10x=-25\Rightarrow {}^{}-10x+$

1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)

\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)

\(=\dfrac{-1}{2}+\dfrac{4}{5}\)

\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)

2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)

\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)

\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)

3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)

\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)

\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)

\(=\dfrac{17}{7}\)

4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)

\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)

\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)

\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)

\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)

\(x^3+1\)

\(=x^3+1^3\)

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

______

\(8+x^3\)

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

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

______

\(27x^3-64y^3\)

\(=\left(3x\right)^3-\left(4y\right)^3\)

\(=\left(3x-4y\right)\left(9x^2+12xy+16y^2\right)\)

______

\(\dfrac{x^3}{64}-\dfrac{1}{125}\)

\(=\left(\dfrac{x}{4}\right)^3-\left(\dfrac{1}{5}\right)^3\)

\(=\left(\dfrac{x}{4}-\dfrac{1}{5}\right)\left(\dfrac{x^2}{16}+\dfrac{x}{20}+\dfrac{1}{25}\right)\)

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

x^3+8=(x+2)(x^2-2x+4)

27x^3-64y^3=(3x-4y)(9x^2+12xy+16y^2)

\(\dfrac{x^3}{64}-\dfrac{1}{25}=\left(\dfrac{1}{4}x-\sqrt[3]{\dfrac{1}{5}}\right)\left(\dfrac{1}{16}x^2+\dfrac{1}{4\sqrt[3]{5}}\cdot x+\dfrac{1}{\sqrt[3]{25}}\right)\)

x³+1= (x+1)(x²-x+1)

8+x³=2³+x³=(2+x)(4-2x+x²)

27x³-64y³= (3x)³-(4y)³= (3x-4y)(9x²+12xy+16y²)

\(\dfrac{x^3}{64}\)-\(\dfrac{1}{125}\)(sửa đề)=(x/4)³-(1/5)³

=(x/4-1/5)(x/16+x/20+1/25)

\(2\times3\times5+25\times8\times4+70+4\times2\times25\)

\(=15\times2+100\times8+70+100\times2\)

\(=30+800+70+200\)

\(=\left(30+70\right)+\left(800+200\right)\)

\(=100+100\times10\)

\(=100+1000\)

\(=1100\)