\(\left(3x-\frac{5}{12}\right)^2-\frac{121}{64}=0\)

\(\left(3x-\frac{5}{12}\right)^2=\frac{121}{64}\)

\(3x-\frac{5}{12}=\sqrt{\frac{121}{64}}\)

\(3x-\frac{5}{12}=\frac{11}{8}\)

\(3x=\frac{11}{8}+\frac{5}{12}\)

\(3x=\frac{33}{24}+\frac{10}{24}\)

\(3x=\frac{43}{24}\)

\(x=\frac{43}{24}:3\)

\(x=\frac{43}{24}\cdot\frac{1}{3}\)

\(x=\frac{43}{72}\)

\(\left(3x-\dfrac{5}{12}\right)^2-\dfrac{121}{64}=0\)

\(\left(3x-\dfrac{5}{12}\right)^2\) \(=0+\dfrac{121}{64}\)

\(3x-\dfrac{5}{12}\) \(=\sqrt{\dfrac{121}{64}}=\dfrac{11}{8}\)

\(3x\) \(=\dfrac{11}{8}+\dfrac{5}{12}=\dfrac{33+10}{24}=\dfrac{43}{24}\)

\(x\) \(\dfrac{3.43}{24}=\dfrac{43}{8}\)

a) x – 32 : 16 = 48 ó x – 2 = 48 ó x = 48 + 2 ó x = 50

b) 88 – 3.(7+x) = 64 ó 3.(7+x) = 88 – 64 ó 7 + x = 24:3 ó x = 8 – 7 ó x = 1

c) (5+4x) : 3 – 121 : 11 = 4 ó (5+4x) : 3 – 11 = 4 ó (5+4x) : 3 = 4 + 11 ó 5+4x = 15.3 ó 4x = 45 – 5 ó 4x = 40 ó x = 10

d) 15 – 2(3x+1) = 11.13 – 130 ó 15 – 2(3x+1) = 143 – 130 ó 15 – 2(3x+1) = 13

ó 2(3x+1) = 15 – 13 ó 3x + 1 = 2:2 ó 3x = 1 – 1 ó 3x = 0 ó x = 0

Lời giải:
a. $(x^2-9)(5x+15)=0$

$\Rightarrow x^2-9=0$ hoặc $5x+15=0$
Nếu $x^2-9=0$

$\Rightarrow x^2=9=3^2=(-3)^2$

$\Rightarrow x=3$ hoặc $-3$
Nếu $5x+15=0$

$\Rightarrow x=-3$
b.

$x^2-8x=0$
$\Rightarrow x(x-8)=0$

$\Rightarrow x=0$ hoặc $x-8=0$

$\Rightarrow x=0$ hoặc $x=8$

c. 

$5+12(x-1)^2=53$

$12(x-1)^2=53-5=48$

$(x-1)^2=48:12=4=2^2=(-2)^2$

$\Rightarrow x-1=2$ hoặc $x-2=-2$
$\Rightarrow x=3$ hoặc $x=0$

d.

$(x-5)^2=36=6^2=(-6)^2$
$\Rightarrow x-5=6$ hoặc $x-5=-6$

$\Rightarrow x=11$ hoặc $x=-1$

e.

$(3x-5)^3=64=4^3$

$\Rightarrow 3x-5=4$

$\Rightarrow 3x=9$

$\Rightarrow x=3$

f.

$4^{2x}+2^{4x+3}=144$
$2^{4x}+2^{4x}.8=144$

$2^{4x}(1+8)=144$

$2^{4x}.9=144$

$2^{4x}=144:9=16=2^4$

$\Rightarrow 4x=4\Rightarrow x=1$

(x-1)⁵=1

(x-1)⁵=1⁵

x-1=1

x=1+1

x=2

(3x-1)²=121

(3x-1)²=11²

3x-1=11

3x=11+1

3x=12

x=12:3

x=4

4^2x+1=64

4^2x+1=4³

=>2x+1=3

=>2x=3-1

=>2x=2

=>x=2:2

=>x=1

(x-2)⁵=32

(x-2)⁵=2⁵

x-2=2

x=2+2

x=4

\(a.\)

\(\Leftrightarrow x-5+x-7=-12.\)

\(\Leftrightarrow x=-12+-12\)

\(\Leftrightarrow x=-24\)

\(b.\)

\(\Leftrightarrow3x+2x=-109+44\)

\(\Leftrightarrow5x=-65\)

\(\Rightarrow x=-13\)

\(c.\)

\(\Leftrightarrow2x-10-15+5x=-109\)

\(\Leftrightarrow2x-5x=-109+15+10\)

\(\Leftrightarrow3x=-84\)

\(\Rightarrow x=-28\)

\(d.\)

Mik ko biết làm dạng này!Xin lỗi bạn nhé!

.............  Hok Tốt nhé  ..............

........  Nhớ k cho mik nhé .........

.............  Thỏ Ruby  .............

d) \(\left(x-3\right)\left(x+5\right)< 0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3< 0\\x+5< 0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x< 3\\x< -5\end{cases}}\)

ko xem dc de ban oi :/

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