Mình làm 5 câu thôi nhé !:
1) \(7^x\cdot49=7^{90}\)

\(\Rightarrow7^x\cdot7^2=7^{90}\)

\(\Rightarrow7^{x+2}=7^{90}\)

\(\Rightarrow x=90-2\)

\(\Rightarrow x=88\)

2) \(2^x\cdot4=128\)

\(\Rightarrow2^x\cdot2^2=2^7\)

\(\Rightarrow2^{x+2}=2^7\)

\(\Rightarrow x=7-2\)

\(\Rightarrow x=5\)

3) \(3^x-1=2^4\cdot5\)

\(\Rightarrow3^x=80+1\)

\(\Rightarrow3^x=81\)

\(\Rightarrow3^x=3^4\)

\(\Rightarrow x=4\)

4) \(3^x+15=42\)

\(\Rightarrow3^x=42-15\)

\(\Rightarrow3^x=27\)

\(\Rightarrow3^x=3^3\)

\(\Rightarrow x=3\)

5) \(4\cdot2^x-3=125\)

\(\Rightarrow2^2\cdot2^x=128\)

\(\Rightarrow2^{x+2}=2^7\)

\(\Rightarrow x=7-2\)

\(\Rightarrow x=5\)

6) 

\(5^x=5^{15}:5^3\\ \Leftrightarrow5^x=5^{15-3}\\ \Leftrightarrow5^x=5^{12}\\ \Leftrightarrow x=12\)

7) 

\(4^x=4^{15}:16\\ \Leftrightarrow4^x=4^{15}:4^2\\ \Leftrightarrow4^x=4^{15-2}\\ 4^x=4^{13}\\ \Leftrightarrow x=13\)

8) 

\(7^x=7^{20}:7^{10}\\ \Leftrightarrow7^x=7^{20-10}\\ \Leftrightarrow7^x=7^{10}\\ \Leftrightarrow x=10\)

9) 

\(11^x=11^{11}:11\\ \Leftrightarrow11^x=11^{11-1}\\ \Leftrightarrow11^x=11^{10}\\ \Leftrightarrow x=10\)

10)

\(3^{15}.3^x=3^{30}\\ \Leftrightarrow3^x=3^{30}:3^{15}\\ 3^x=3^{30-15}\\ \Leftrightarrow3^x=3^{15}\\ \Leftrightarrow x=15\)

11)

\(10+2x=4^5:4^3\\ \Leftrightarrow10+2x=4^2\\ \Leftrightarrow2x=6\\ \Leftrightarrow x=3\)

12)

\(20+5x=5^5:5^3\\ \Leftrightarrow20+5x=5^2\\ \Leftrightarrow5x=5\\ \Leftrightarrow x=1\)

13)

\(4x-20=2^5:2^2\\ \Leftrightarrow4x-20=2^3\\ \Leftrightarrow4x=28\\ \Leftrightarrow x=12\)

14)

\(15+2x=3^{80}:3^{77}\\ \Leftrightarrow15+2x=3^3\\ \Leftrightarrow2x=12\\ \Leftrightarrow x=6\)

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0 

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52 

=> 19x = -4

=> x = -4/19

d/ 20x² - 16x - 34 = 10x² + 3x - 34

=> 10x² - 19x = 0

=> x(10x - 19) = 0

=> x = 0 

hoặc 10x - 19 = 0 => 10x = 19 => x = 19/10

Vậy x = 0 ; x = 19/10

Rút gọn hết ta được :

a/ 41x - 17 = -21

=> 41x = -4 => x = 4/41

b/ 34x - 17 = 0

=> 34x = 17

=> x = 17/34 = 1/2

c/ 19x + 56 = 52

=> 19x = -4

=> x = -4/19

d/ 20x 2 - 16x - 34 = 10x 2 + 3x - 34

=> 10x 2 - 19x = 0

=> x(10x - 19) = 0

=> x = 0 hoặc 10x - 19 = 0

=> 10x = 19

=> x = 19/10

Vậy x = 0 ; x = 19/10 

a) ( 6x - 3 ) ( 2x + 4 ) + ( 4x - 1 ) ( 5 - 3x ) = -21

<=> 12x² + 24x - 6x - 12 + 20x - 12x² - 5 + 3x = -21

<=> 41x = -21 + 12 + 5 

<=> 41x = -4

<=> x = -4/41

a) \(2x^2+20x+52=0\Rightarrow x^2+10x+26=0\Rightarrow\left(x+5\right)^2+1=0\)

\(\Rightarrow\) vô nghiệm

b) ĐK: \(x\ne1;-1\)

\(\dfrac{2x-19}{5x^2-5}-\dfrac{17}{x-1}=\dfrac{8}{1-x}\Rightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}-\dfrac{17}{x-1}+\dfrac{8}{x-1}=0\)

\(\Rightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}-\dfrac{9}{x-1}=0\Rightarrow\dfrac{2x-19-45\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}=0\)

\(\Rightarrow-43x-64=0\Rightarrow x=-\dfrac{64}{43}\)

a)  Ta có: \(\Delta'=100-104=-4< 0\)

Vậy phương trình vô nghiệm.

b) ĐKXĐ: \(x\ne1;x\ne-1\)

\(\Leftrightarrow\dfrac{2x-19}{5\left(x^2-1\right)}=\dfrac{17}{x-1}-\dfrac{8}{x-1}\)

\(\Leftrightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}=\dfrac{9}{x-1}\)

\(\Leftrightarrow\dfrac{2x-19}{5\left(x-1\right)\left(x+1\right)}=\dfrac{45\left(x+1\right)}{5\left(x-1\right)\left(x+1\right)}\)

\(\Rightarrow2x-19=45x+45\)

\(\Leftrightarrow43x=-64\)

\(\Leftrightarrow x=-\dfrac{64}{43}\)(TM)

Vậy phương trình có nghiệm là: \(x=-\dfrac{64}{43}\)

Bài 1:

a) \(6x\left(3x+15\right)-2x\left(9x-2\right)=17\) (1)

\(\Leftrightarrow18x^2+90x-18x^2+4x=17\)

\(\Leftrightarrow94x=17\)

\(\Leftrightarrow x=\dfrac{17}{94}\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{17}{94}\right\}\)

b) \(\left(15x-2x\right)\left(4x+1\right)-\left(13x-4x\right)\left(2x-3\right)-\left(x-1\right)\left(x+2\right)+x+2=52\)

\(\Leftrightarrow\left(60x^2+15x-8x^2-2x\right)-\left(26x^2-39x-8x^2+12x\right)-\left(x^2+2x-x-2\right)+x+2=52\)

\(\Leftrightarrow60x^2+15x-8x^2-2x-26x^2+39x+8x^2-12x-x^2-2x+x+2+x+2=52\)

\(\Leftrightarrow33x^2+40x+4=52\)

\(\Leftrightarrow33x^2+40x=48\)

...

Bài 1 có ng làm rồi nên mình không làm nx nhé.

2) a) Rút gọn

P=\(3x\left(4x+1\right)+5x^2-4x\left(3x+9\right)+x\left(5x-5x^2\right)\)

P= \(12x^2+3x+5x^3-12x^3-36x+5x^2-5x^3\)

P= \(-33x\)

b) |x| = 2

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

Với x = 2 \(\Rightarrow\) P = -33 . 2 = -66

Với x = -2 \(\Rightarrow\) P = -33 . (-2) = 66

c) Để P = 2017 \(\Rightarrow\) -33x = 2017 \(\Rightarrow\) x = \(-\dfrac{2017}{33}\)

Bài 3: Giải

f(x) = \(\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

f(x) = \(\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

f(x) = \(\left(x^2+5x\right)^2-6^2\) ( Hằng đẳng thức số 3 )

f(x) = \(\left(x^2+5x\right)^2-36\ge-36\) với mọi x

Vậy \(Min_{f\left(x\right)}\) = -36 khi x = 0 hoặc x = -5

Nguyễn Huy Tú tl dùm đc k mk cần lắm

Thanks nhiều

   x²-4x+5=0

=>(x-2)²+1=0

=>(x-2)² =-1

=> pt vô nghiệm

   (x²+5x)(x³+3x²-18x)=0

=>\(\int^{x^2+5x=0}_{x^3+3x^2-18x=0}=>\int^{\int^{x=0}_{x=-5}}_{x=3;x=0;x=-6}\)

\(5x\left(4x^2-2x+1\right)-2x\left(10^2-5x-2\right)\)

\(=5x.4x^2-5x.2x+5x.1\)

5x(4x²-2x+1)-2x(10²-5x-2) 

=20X³-10X²+5X-196X+10X²

=20X³-(-10X²+10X²)+(5X-196X)

=20X³-191X

Tới đây thay x và giải tiếp

hình như đề sai ở chỗ 10^2 fai là 10x^2 thì rút gọn số ms bé chứ nếu=10^2 khi rút xong và thay số to lắm

5x(4x^2-2x+1)-2x(10^2-5x-2)

câu f là 9+3x hay 9-3x vậy???

a) 

$2x+6=0$

$2x=-6$

$x=-3$

b) $4x+20=0$

$4x=-20$

$x=-5$

c) 

$2(x-1)=5x-7$

$2x-2=5x-7$

$3x=5$

$x=\frac{5}{3}$

d) $2x-3=0$

$2x=3$

$x=\frac{3}{2}$

e) 

$3x-1=x+3$

$2x=4$

$x=2$

f) 

$15-7x=9-3x$

$6=4x$

$x=\frac{3}{2}$

g) $x-3=18$

$x=18+3=21$

h) 

$2x+1=15-5x$

$7x=14$

$x=2$

\(e,112-45+5x=87\)

\(67+5x=87\)

\(5x=20\)

\(x=4\)

\(f,6^2+64:\left(x-1\right)=52\)

\(36+64:\left(x+1\right)=52\)

\(64:\left(x+1\right)=16\)

\(x+1=4\)

\(x=3\)

\(g,\left(55-x\right)\left(2x-44\right)=0\)

\(\Rightarrow\hept{\begin{cases}55-x=0\\2x-44=0\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=55\\x=22\end{cases}}\)

Vậy ....

\(h,x^2+4=13\)

\(x^2=9\)

\(x^2=\pm3^2\)

\(\Rightarrow x=\pm3\)

\(i,2\left(x+1\right)^3=16\)

\(\left(x+1\right)^3=8\)

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

\(\Rightarrow x+1=2\)

\(x=1\)

\(g,150-5\left(x-2\right)^2=25\)

\(5\left(x-2\right)^2=125\)

\(\left(x-2\right)^2=25\)

\(\left(x-2\right)^2=\pm5^2\)

\(\Rightarrow\hept{\begin{cases}x-2=5\\x-2=\left(-5\right)\end{cases}\Rightarrow\hept{\begin{cases}x=7\\x=\left(-3\right)\end{cases}}}\)

Vậy ....