Bài 4:
a) \(2\left(3x-1\right)=10\)
\(\Rightarrow3x-1=\dfrac{10}{2}\)
\(\Rightarrow3x-1=5\)
\(\Rightarrow3x=5+1\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=\dfrac{6}{3}\)
\(\Rightarrow x=2\)
b) \(\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=49\)
\(\Rightarrow\left(9x^2+24x+16\right)-\left(9x^2-1\right)=49\)
\(\Rightarrow9x^2+24x+16-9x^2+1=49\)
\(\Rightarrow24x+16=49-1\)
\(\Rightarrow24x+16=48\)
\(\Rightarrow24x=48-16\)
\(\Rightarrow24x=32\)
\(\Rightarrow x=\dfrac{32}{24}\)
\(\Rightarrow x=\dfrac{4}{3}\)
4:
a: 2(3x-1)=10
=>3x-1=5
=>3x=6
=>x=2
b: \(\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=49\)
=>\(9x^2+24x+16-9x^2+1=49\)
=>\(24x+17=49\)
=>24x=32
=>\(x=\dfrac{4}{3}\)
3:
a: \(\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2\)
\(=\left(x+8-x+2\right)^2=10^2=100\)
b: \(\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x^3+1\right)-\left(x^3-1\right)\)
\(=x^3+1-x^3+1\)
=2
Câu 4:
a)
`2(3x-1)=10`
<=> `3x-1=5`
<=> `3x=6`
<=> `x=2`
b)
<=> `9x^2+24x+16-9x^2+1=49`
<=> `24x=49-1-16=32`
<=> `x=(32)/(24)`
<=> `x=(4)/(3)`
Câu 3.
\(a,\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2\)
\(=\left[\left(x+8\right)-\left(x-2\right)\right]^2\)
\(=\left(x+8-x+2\right)^2\)
\(=10^2=100\)
⇒ Giá trị của biểu thức không phụ thuộc vào giá trị của biến.
\(b,\left(x+1\right)\left(x^2-x+1\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=\left(x^3+1^3\right)-\left(x^3-1^3\right)\)
\(=x^3+1-x^3+1\)
\(=2\)
⇒ Giá trị của biểu thức không phụ thuộc vào giá trị của biến.
Câu 4.
\(a,2\left(3x-1\right)=10\)
\(\Rightarrow3x-1=10:2\)
\(\Rightarrow3x-1=5\)
\(\Rightarrow3x=5+1\)
\(\Rightarrow3x=6\)
\(\Rightarrow x=6:3=2\)
Vậy: \(x=2.\)
\(b,\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=49\)
\(\Leftrightarrow\left(3x\right)^2+2\cdot3x\cdot4+4^2-\left[\left(3x\right)^2-1^2\right]=49\)
\(\Leftrightarrow9x^2+24x+16-\left(9x^2-1\right)=49\)
\(\Leftrightarrow9x^2+24x+16-9x^2+1=49\)
\(\Leftrightarrow24x+17=49\)
\(\Leftrightarrow24x=49-17\)
\(\Leftrightarrow24x=32\)
\(\Leftrightarrow x=\dfrac{32}{24}=\dfrac{4}{3}\)
Vậy: \(x=\dfrac{4}{3}.\)
\(Toru\)
