x=3

b,Dat an 2x^2-3x-1=a la dc

a, \(4^x-10.2^x+16=0\Leftrightarrow\left(2^x\right)^2-10.2^x+16=0\)

Đặt \(2^x=t\Rightarrow t^2-10t+16=0\Leftrightarrow\orbr{\begin{cases}t=8\\t=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)

b. Đặt \(2x^2-3x-1=t\Rightarrow t^2-3\left(t-4\right)-16=0\)

\(\Leftrightarrow t^2-3t-28=0\Leftrightarrow\orbr{\begin{cases}t=7\\t=-4\end{cases}}\)

Thế vào rồi giải tiếp em nhé.

camon ạ

Câu 1 :

a. \(4x-5=23\\ \Leftrightarrow4x=23+5\\ \Leftrightarrow4x=28\\ \Leftrightarrow x=7\)

b. 

|-2x|=5x+14

 Nếu - 2x > 0 => x < 0 thì |-2x|= - 2x, ta có pt: -2x = 5x+14

 <=> - 2x = 5x + 14

 <=> - 2x - 5x = 14

 <=> - 7x = 14

 <=> x = - 2 (thoã mãn)

 Nếu - 2x < 0 => x > 0 thì |-2x|= = -(- 2x) = 2x.

Ta có pt: 2x = 5x + 14

 <=> - 3x = 14

<=> x = \(-\dfrac{14}{3}\)
 Vậy pt có nghiệm x = - 2

c) \(\dfrac{x+1}{x-1}-\dfrac{1}{x+1}=\dfrac{x^2+2}{x^2-1}\\ ĐKXĐ:x\ne1;x\ne-1\\ \Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{1\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2+2}{\left(x-1\right)\left(x+1\right)}\\ \Leftrightarrow x^2+x+x+1-x+1=x^2+2\\ \Leftrightarrow x^2+x+x-x-x^2=2-1-1\\ \Leftrightarrow x=0\left(nhận\right)\)

\(a,4x-5=23\)

\(\Leftrightarrow4x=23+5\)

\(\Leftrightarrow4x=28\)

\(\Leftrightarrow x=7\)

\(b,\left|-2x\right|=5x+14\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=5x+14\\2x=-5x-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x-14=0\\7x+14=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-3x=14\\7x=-14\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{14}{3}\\x=-2\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{14}{3};-2\right\}\)

\(c,\Leftrightarrow\dfrac{\left(x+1\right)\left(x+1\right)-x+1-x^2-2}{x^2-1}=0\)

\(\Leftrightarrow x^2+x+x+1-x+1-x^2-2=0\)

\(\Leftrightarrow x=0\)

Vậy \(S=\left\{0\right\}\)

a) \(4x-5=23\)

    \(4x=23+5\)

      \(4x=28\)

        \(x=7\)

b) \(\left|-2x\right|=5x+14\)

 \(\Leftrightarrow\)   \(-2x-5=14\)

\(\Leftrightarrow\)    \(-7x=14\)

\(\Leftrightarrow\)         \(x=-2\)

\(\Leftrightarrow\)    \(-2x=-\left(5x+14\right)\)

\(\Leftrightarrow\)    \(-2x=-\left(5x-14\right)\)

\(\Leftrightarrow\)  \(-2x+5x=-14\)

 \(\Leftrightarrow\)    \(3x=-14\)

 \(\Leftrightarrow\) \(x=-\dfrac{14}{3}\) \(\left(\text{vô lí}\right)\)

  \(\Leftrightarrow x=-2\)   

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

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

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

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

\(\Leftrightarrow x+2=2\)

\(\Leftrightarrow x=0\)

2x + 13/6 =8/27

2x            = 8/27 - 13/6

2x            = - 101/54

x            = - 101/54 : 2

x              = - 101/108

\(\left|2x-\frac{1}{2}\right|+1=3x\)

\(\Leftrightarrow\left|2x-\frac{1}{2}\right|=3x-1\)

\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{1}{2}=3x-1\\2x-\frac{1}{2}=1-3x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-3x=-1+\frac{1}{2}\\2x+3x=1+\frac{1}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x=-\frac{1}{2}\\5x=\frac{3}{2}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{3}{10}\end{cases}}\)

a,\(\left(1+x\right)^3=\left(2x\right)^3\)

=>\(1+x=2x\)

=>\(x-2x=-1\)

=>\(-x=-1\)

=>\(x=1\)

vậy ​\(x=1\)​

b,\(\left(x-1\right)^2=16\)

=>\(\left(x-1\right)^2=4^2\)

=>\(x-1=4\)

=>\(x=4+1\)

=>\(x=5\)

Vậy​\(x=5\)​

c,\(\left(x+1\right)^2=25\)

=>\(\left(x+1\right)^2=5^2\)

=>\(x+1=5\)

=>\(x=5-1\)

=>\(x=4\)

Vậy \(x=4\)

d,\(4x^3+15=47\)

=>\(4x^3=47-15\)

=>\(4x^3=32\)

=>\(x^3=32:4\)

=>\(x^3=8\)

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

=>\(x=2\)

Vậy\(x=2\)

e,\(\left(2x-1\right)^5=x^5\)

=>\(2x-1=x\)

=>\(2x-x=1\)

=>\(x=1\)

Vậy\(x=1\)

ĐÚNG K MÌNH NHA

a.x(y+3)=3

=> x(y+3) ∈Ư(3)={-3;-1;1;3}

ta có bảng sau

vậy x=-3 thì y=-4

x=-1 thì y=-6

x=1 thì y=0

x=3 thì y=-2

c.x+3⋮ x+1

=> (x+3)-(x+1)⋮(x+1)

=> (x+3-x-1)⋮(x+1)

=> 2⋮(x+1)

=> (x+1) ∈ Ư(2)={-2;-1;1;2}

=> x∈{-3;-2;0;1}

vậy x ∈{-3;-2;0;1}

b,d tương tự

a.(x-2)(x+3)>0

=>\(\left[{}\begin{matrix}x-2>0\\x+3>0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>2\\x>-3\end{matrix}\right.\)

=> x>2

vậy x>2

b.(x-2)(x-1)>0

=> \(\left[{}\begin{matrix}x-2>0\\x-1>0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>2\\x>1\end{matrix}\right.\)

=> x>2

vậy x>2

c.(x-2)(x²+1)>0

=> \(\left[{}\begin{matrix}x-2>0\\x^2+1>0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>2\\x^2>-1\Rightarrow x>\sqrt{-1}\end{matrix}\right.\)

vậy x>2

d.(x-1)(x+2)>0

=> \(\left[{}\begin{matrix}x-1>0\\x+2>0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>1\\x>-2\end{matrix}\right.\)

=> x>1

vậy x>1

Còn câu này nx bn ạ:

x^2.(x+2)<0

Tìm x

Giúp mk nhanh nha, mk cần gấp