a) \(3^{x+1}-3^x=162\)

\(\Leftrightarrow3^x.\left(3-1\right)=162\)

\(\Leftrightarrow3^x.2=162\)

\(\Leftrightarrow3^x=162:2=81\)

\(\Leftrightarrow3^x=3^4\)

\(\Leftrightarrow x=4\)

b) \(\left(1-x\right)^3=216\)

\(\Leftrightarrow\left(1-x\right)^3=6^3\)

\(\Leftrightarrow1-x=6\)

\(\Leftrightarrow x=1-6\)

\(\Leftrightarrow x=-5\)

c) \(5^{x+1}-2.5^x=375\)

\(\Leftrightarrow5^x.\left(5-2\right)=375\)

\(\Leftrightarrow5^x.3=375\)

\(\Leftrightarrow5^x=375:3=125\)

\(\Leftrightarrow5^x=5^3\)

\(\Leftrightarrow x=3\)

d) \(\left(2x+1\right)^3=-64\)

=> \(\left(2x+1\right)^3=\left(-4\right)^3\)

=> \(2x+1=-4\)

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

=> \(2x=-5\)

=> \(x=\left(-5\right):2\)

=> \(x=-\frac{5}{2}\)

Vậy \(x=-\frac{5}{2}.\)

Mình chỉ làm câu d) thôi nhé.

Chúc bạn học tốt!

Gửi bạn Velvet Red ! Hơi bận nên chỉ làm hai phần a,b cho bạn thôi nhé !!

a) \(3^{x+2}-3^{x+1}=162\)

\(\Leftrightarrow3^x.3^2-3^x.3=162\)

\(\Leftrightarrow3^x.\left(3^2-3\right)=162\)

\(\Leftrightarrow3^x=162:6\)

\(\Leftrightarrow3^x=27=3^3\)

\(\Leftrightarrow x=3\)

Vậy : \(x=3\)

b) \(5^{x+2}-2.5^{x+1}=375\)

\(\Leftrightarrow5^x.5^2-2.5^x.5=375\)

\(\Leftrightarrow5^x.\left(5^2-2.5\right)=375\)

\(\Leftrightarrow5^x=375:15\)

\(\Leftrightarrow5^x=25=5^2\)

\(\Leftrightarrow x=2\)

Vậy : \(x=2\)

\(a)3^{x+1}-3^x=162\)

\(\Leftrightarrow3^x\cdot3-3^x=162\)

\(\Leftrightarrow3^x\left(3-1\right)=162\)

\(\Leftrightarrow3x\cdot2=162\)

\(\Leftrightarrow3x=162:2\)

\(\Leftrightarrow3x=81\)

\(\Leftrightarrow x=81:3\)

\(\Leftrightarrow x=27\)

Vậy x=27

\(b)\left(1-x\right)^3=216\)

\(\Leftrightarrow\left(1-x\right)^3=6^3\)

\(\Leftrightarrow x-1=6\)

\(\Leftrightarrow x=6+1\)

\(\Leftrightarrow x=7\)

Vậy x=7

\(c)5^{x+1}-2\cdot5^x=375\)

\(\Leftrightarrow5^x\cdot5-2\cdot5^x=375\)

\(\Leftrightarrow5^x\cdot\left(5-2\right)=375\)

\(\Leftrightarrow5^x\cdot3=375\)

\(\Leftrightarrow5^x=375:3\)

\(\Leftrightarrow5^x=125\)

\(\Leftrightarrow5^x=5^3\)

\(\Leftrightarrow x=3\)

Vậy x=3

(3x+1)^3=-216=(-6)^3

=>3x+1=-6

=>x=...

3^x-1+5.3^x-1=162

=>3^x-1.(1+5)=162

=>3^x-1.6=162

=>3^x-1=162:6=27

=>3^x-1=3^3

=>x-1=3

=>x=4

a/ \(x=\dfrac{-5}{12}\)

b/ \(x\approx-1,9526\)

c/ \(x=\dfrac{21-i\sqrt{199}}{10}\)

d/ \(x=\dfrac{-20}{13}\)

a) (x-2)³+6(x+1)²-x³+12=0

⇒ x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

⇒ x³-6x²+12x-8+6x²+12x+6-x³+12=0

⇒ 24x+10=0

⇒ 24x=-10

⇒ x=-5/12

a.

PT \(\Leftrightarrow x^3-6x^2+12x-8+6(x^2+2x+1)-x^3+12=0\)

\(\Leftrightarrow x^3-6x^2+12x-8+6x^2+12x+6-x^3+12=0\)

\(\Leftrightarrow 24x+10=0\Leftrightarrow x=\frac{-5}{12}\)

b. Bạn xem lại đề, nghiệm khá xấu không phù hợp với mức độ tổng thể của bài.

c.

PT $\Leftrightarrow (4x^2+12x+9)+(x^2-1)=5(x^2+4x+4)+(x^2-4x-5)+9(x^2+6x+9)$
$\Leftrightarrow 10x^2+42x+64=0$

$\Leftrightarrow x^2+(3x+7)^2=-15< 0$ (vô lý) 

Do đó pt vô nghiệm.

d.

PT $\Leftrightarrow (1-6x+9x^2)-(9x^2-17x-2)=(9x^2-16)-9(x^2+6x+9)$

$\Leftrightarrow 11x+3=-54x-97$

$\Leftrightarrow 65x=-100$

$\Leftrightarrow x=\frac{-20}{13}$

1a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=1-4x\end{cases}}\)

=> \(\orbr{\begin{cases}-\frac{5}{2}x=-\frac{3}{2}\\\frac{11}{2}x=\frac{1}{2}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{5}{3}\\x=\frac{1}{11}\end{cases}}\)

b) \(\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=>\(\left|\frac{5}{4}x-\frac{7}{2}\right|=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\orbr{\begin{cases}\frac{5}{4}x-\frac{7}{2}=\frac{5}{8}x+\frac{3}{5}\\\frac{5}{4}x-\frac{7}{2}=-\frac{5}{8}x-\frac{3}{5}\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{5}{8}x=\frac{41}{10}\\\frac{15}{8}x=\frac{29}{10}\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c) TT

a, \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\-\frac{3}{2}x-\frac{1}{2}=4x-1\end{cases}}\)

=> \(\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}-4x=-1\\-\frac{3}{2}x-\frac{1}{2}-4x=-1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\)

\(b,\left|\frac{5}{4}x-\frac{7}{2}\right|-\left|\frac{5}{8}x+\frac{3}{5}\right|=0\)

=> \(\left|\frac{5}{4}x-\frac{7}{2}\right|-0=\left|\frac{5}{8}x+\frac{3}{5}\right|\)

=> \(\frac{\left|5x-14\right|}{4}=\frac{\left|25x+24\right|}{40}\)

=> \(\frac{10(\left|5x-14\right|)}{40}=\frac{\left|25x+24\right|}{40}\)

=> \(\left|50x-140\right|=\left|25x+24\right|\)

=> \(\orbr{\begin{cases}50x-140=25x+24\\-50x+140=25x+24\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{164}{25}\\x=\frac{116}{75}\end{cases}}\)

c, \(\left|\frac{7}{5}x+\frac{2}{3}\right|=\left|\frac{4}{3}x-\frac{1}{4}\right|\)

=> \(\orbr{\begin{cases}\frac{7}{5}x+\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\\-\frac{7}{5}x-\frac{2}{3}=\frac{4}{3}x-\frac{1}{4}\end{cases}}\)

=> \(\orbr{\begin{cases}x=-\frac{55}{4}\\x=-\frac{25}{164}\end{cases}}\)

Bài 2 : a. |2x - 5| = x + 1

 TH1 : 2x - 5 = x + 1

    => 2x - 5 - x = 1

    => 2x - x - 5 = 1

    => 2x - x = 6

    => x = 6

TH2 : -2x + 5 = x + 1

   => -2x + 5 - x = 1

   => -2x - x + 5 = 1

   => -3x = -4

   => x = 4/3

Ba bài còn lại tương tự

cho mình

a: \(A\left(x\right)+B\left(x\right)\)

\(=-2x^3+11x^2-5x-\dfrac{1}{5}+2x^3-3x^2-7x+\dfrac{1}{5}\)

\(=8x^2-12x\)

b: C(x)=A(x)-B(x)

\(=-2x^3+11x^2-5x-\dfrac{1}{5}-2x^3+3x^2+7x-\dfrac{1}{5}\)

\(=-4x^3+14x^2+2x-\dfrac{2}{5}\)

a) 3^x-1(1+5)=162

3^x-1.6=162

3^x-1=162:6=27=3³

=>x-1=3

x=4

b) x(x+3)=0

=>x=0 hoặc x+3=0

=>x=0 hoặc x=-3

c) Vì tích nhỏ hơn 0 nên có 1 thừa số dương và 1 thừa số âm

Có x-1>x-3

=>x-1>0 và x-3<0

=>x>1 và x<3

Vậy x=2

a) 3^x-1 + 5. 3^x-1 = 162

  1. 3^x-1 + 5. 3^x-1  = 162

( 1 + 5 ) . 3^x-1 = 162

6. 3^x-1 = 162

    3^x-1 = 162 : 6

    3^x-1 = 27

    3^x-1 = 3³

    x - 1 =3

    x       = 3 + 1

    x        = 4

b) 3x + x² = 0

    3x + x.x = 0

    3x + 1x + 1x =0

   ( 3 + 1 + 1 ).x = 0

      5x    = 0

        x = 0 : 5

        x = 0

a) Ta có: \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)

\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)

\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x+25=15\)

\(\Leftrightarrow24x=-10\)

hay \(x=-\dfrac{5}{12}\)

b) Ta có: \(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x-5\right)\left(x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)

c) Ta có: \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\)

d) Ta có: \(x^3-x=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)=0\)

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

e) Ta có: \(27x^3-27x^2+9x-1=1\)

\(\Leftrightarrow\left(3x\right)^3-3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2-1^3=1\)

\(\Leftrightarrow\left(3x-1\right)^3=1\)

\(\Leftrightarrow3x-1=1\)

\(\Leftrightarrow3x=2\)

hay \(x=\dfrac{2}{3}\)