(1/2)x-1 + 3 = 25/8
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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

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

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
a) 7x(x+1)−3(x+1)=0⇒(x+1)(7x−3)=07x(x+1)−3(x+1)=0⇒(x+1)(7x−3)=0
⇒[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


Bài 1.
\(a, (3x-4)^2\)
\(=\left(3x\right)^2-2\cdot3x\cdot4+4^2\)
\(=9x^2-24x+16\)
\(b,\left(1+4x\right)^2\)
\(=1^2+2\cdot1\cdot4x+\left(4x\right)^2\)
\(=16x^2+8x+1\)
\(c,\left(2x+3\right)^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)
\(=8x^3+36x^2+54x+27\)
\(d,\left(5-2x\right)^3\)
\(=5^3-3\cdot5^2\cdot2x+3\cdot5\cdot\left(2x\right)^2-\left(2x\right)^3\)
\(=125-150x+60x^2-8x^3\)
\(e,49x^2-25\)
\(=\left(7x\right)^2-5^2\)
\(=\left(7x-5\right)\left(7x+5\right)\)
\(f,\dfrac{1}{25}-81y^2\)
\(=\left(\dfrac{1}{5}\right)^2-\left(9y\right)^2\)
\(=\left(\dfrac{1}{5}-9y\right)\left(\dfrac{1}{5}+9y\right)\)
Bài 2.
\(a,\left(x-5\right)^2-\left(x+7\right)\left(x-7\right)=8\)
\(\Rightarrow x^2-2\cdot x\cdot5+5^2-\left(x^2-7^2\right)=8\)
\(\Rightarrow x^2-10x+25-\left(x^2-49\right)=8\)
\(\Rightarrow x^2-10x+25-x^2+49=8\)
\(\Rightarrow\left(x^2-x^2\right)-10x=8-25-49\)
\(\Rightarrow-10x=-66\)
\(\Rightarrow x=\dfrac{33}{5}\)
\(b,\left(2x+5\right)^2-4\left(x+1\right)\left(x-1\right)=10\)
\(\Rightarrow\left(2x\right)^2+2\cdot2x\cdot5+5^2-4\left(x^2-1^2\right)=10\)
\(\Rightarrow4x^2+20x+25-4x^2+4=10\)
\(\Rightarrow\left(4x^2-4x^2\right)+20x=10-25-4\)
\(\Rightarrow20x=-19\)
\(\Rightarrow x=\dfrac{-19}{20}\)
#\(Toru\)
Bài 1
a) (3x - 4)²
= (3x)² - 2.3x.4 + 4²
= 9x² - 24x + 16
b) (1 + 4x)²
= 1² + 2.1.4x + (4x)²
= 1 + 8x + 16x²
c) (2x + 3)³
= (2x)³ + 3.(2x)².3 + 3.2x.3² + 3³
= 8x³ + 36x² + 54x + 27
d) (5 - 2x)³
= 5³ - 3.5².2x + 3.5.(2x)² - (2x)³
= 125 - 150x + 60x² - 8x³
e) 49x² - 25
= (7x)² - 5²
= (7x - 5)(7x + 5)
f) 1/25 - 81y²
= (1/5)² - (9y)²
= (1/5 - 9y)(1/5 + 9y)

a)\(x-\frac{1}{4}=\frac{5}{8}\)
\(x=\frac{5}{8}+\frac{1}{4}\)
\(x=\frac{7}{8}\)
b)\(\frac{x}{12}=\frac{-1}{24}-\frac{1}{8}\)
\(\frac{x}{12}=\frac{-1}{24}+\frac{-1}{8}\)
\(\frac{x}{12}=\frac{-1}{6}\)
\(x=\frac{-1}{6}\times12\)
\(x=-2\)
a)x-1/4=5/8
=>x=5/8+1/4
x=7/8
b) x/12=-1/24-1/8
=>x/12=-1/6
=>x=-2
c).......... Tự làm

\(a,121-\left(115+x\right)=3x-\left(25-9-5x\right)-8\\ 121-115-x=3x-25+9+5x-8\\ 6-x=8x-24\\ 8x+x=-24-6\\ 9x=-30\\ x=-\dfrac{30}{9}=-\dfrac{10}{3}\\ ----\\ b,2^{x+2}.3^{x+1}.5^x=10800\\ \left(2.3.5\right)^x.2^2.3=10800\\ 30^x.12=10800\\ 30^x=\dfrac{10800}{12}=900=30^2\\ Vậy:x=2\)
\(\left(\frac12\right)^{x-1}+3=\frac{25}{8}\)
\(\left(\frac12\right)^{x-1}=\frac{25}{8}-3\)
\(\left(\frac12\right)^{x-1}=\frac18=\left(\frac12\right)^3\)
⇒ x - 1 = 3
⇒ x = 3 + 1
⇒ x = 4
vậy x = 4