a) \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)

\(\Leftrightarrow x^2+6x+9-x^2+4x-4=x^3-5x^2+25x+5x^2-25x+125-108\)

\(\Leftrightarrow x^3-10x+12=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+6\right)=0\)

\(\Leftrightarrow x=2\)( do \(x^2+2x+6=\left(x+1\right)^2+4\ge4>0\))

1) \(5^{x+1}-5^x=20\Leftrightarrow5^x\left(5-1\right)=20\Leftrightarrow5^x=5\Leftrightarrow x=1\)

2) \(2^x+2^{x+4}=544\Leftrightarrow2^x\left(1+2^4\right)=544\Leftrightarrow2^x=32\Leftrightarrow x=5\)

3) \(4^{2x+1}+4^{2x}=80\Leftrightarrow4^{2x}\left(4+1\right)=80\Leftrightarrow16^x=16\Leftrightarrow x=1\)

4) \(3^{2x+2}+3^{2x+1}=108\Leftrightarrow3^{2x}\left(3^2+3\right)=108\Leftrightarrow9^x=9\Leftrightarrow x=1\)

5) \(7^{x+3}-7^{x+1}=16464\Leftrightarrow7^x\left(7^3-7\right)=16464\Leftrightarrow7^x=49\Leftrightarrow x=2\)

cảm ơn bn nha,

Ta có: \(4^{2x+4}=4^3\)

\(\Rightarrow\)\(2x+4=3\)

\(\Rightarrow\)\(x=\frac{-1}{2}\)

Ta có: \(3^{x-1}+3^{x-2}=108\)

\(\Rightarrow\)\(3^{x-1}+3^{x-2}=2^2.3^3\)

\(\Rightarrow\)\(3^{x-2}=3^3\)

\(\Rightarrow\)\(x-2=3\)

\(\Rightarrow x=5\)

5: (x-1)(x-5)=(x-1)(x-2)

=>(x-1)(x-5)-(x-1)(x-2)=0

=>(x-1)(x-5-x+2)=0

=>-3(x-1)=0

=>x-1=0

=>x=1

6: \(6\left(x-3\right)\left(x-4\right)-6x\left(x-2\right)=4\)

=>\(6\left(x^2-7x+12\right)-6x^2+12x=4\)

=>\(6x^2-42x+72-6x^2+12x=4\)

=>-30x=4-72=-68

=>\(x=\frac{68}{30}=\frac{34}{15}\)

7: -(x+3)(x-4)+(x+1)(x-1)=10

=>\(-\left(x^2-x-12\right)+x^2-1=10\)

=>\(-x^2+x+12+x^2-1=10\)

=>x+11=10

=>x=-1

8: (2x-1)(x-2)-(2x-7)(x+3)=3

=>\(2x^2-4x-x+2-\left(2x^2+6x-7x-21\right)=3\)

=>\(2x^2-5x+2-\left(2x^2-x-21\right)=3\)

=>\(2x^2-5x+2-2x^2+x+21=3\)

=>-4x+23=3

=>-4x=-20

=>x=5

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

=>\(-\left(x^2-9x+20\right)-\left(x^2+2x-3\right)=-2x^2\)

=>\(-x^2+9x-20-x^2-2x+3=-2x^2\)

=>7x-17=0

=>7x=17

=>x=17/7

10: (4x+1)(x-3)-(x-7)(4x-1)=15

=>\(4x^2-12x+x-3-\left(4x^2-x-28x+7\right)=15\)

=>\(4x^2-11x-3-4x^2+29x-7=15\)

=>18x-10=15

=>18x=25

=>x=25/18

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

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

=>3x=4-1=3

=>x=1

12: \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)

=>\(x^3-27+5x-x^3=6x\)

=>6x=5x-27

=>x=-27

13: (3x-5)(x+1)-(3x-1)(x+1)=x-4

=>(x+1)(3x-5-3x+1)=x-4

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

=>-4x-4=x-4

=>-4x-x=0

=>-5x=0

=>x=0

14; 5(x-3)(x-7)-(5x+1)(x-2)=8

=>\(5\left(x^2-10x+21\right)-\left(5x^2-10x+x-2\right)=8\)

=>\(5x^2-50x+105-5x^2+9x+2=8\)

=>-41x=8-2-105=6-105=-99

=>x=99/41

Lời giải:

a. Áp dụng TCDTSBN:

\(\frac{x}{y}=\frac{2}{5}\Rightarrow \frac{x}{2}=\frac{y}{5}=\frac{2x}{4}=\frac{y}{5}=\frac{2x-y}{4-5}=\frac{3}{-1}=-3\)

$\Rightarrow x=-3.2=-6; y=-3.5=-15$

b. Áp dụng TCDTSBN:

$\frac{x}{2}=\frac{y}{3}; \frac{y}{4}=\frac{z}{7}$

$\Rightarrow \frac{x}{8}=\frac{y}{12}=\frac{z}{21}$

$=\frac{2x}{16}=\frac{y}{12}=\frac{z}{21}=\frac{2x-y+z}{16-12+21}=\frac{50}{25}=2$

$\Rightarrow x=8.2=16; y=2.12=24; z=2.21=42$

c.

$\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$

$\Rightarrow \frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{16}=\frac{2z^2}{32}$

$=\frac{x^2-y^2+2z^2}{4-9+32}=\frac{108}{27}=4$

$\Rightarrow x^2=4.4=16; y^2=9.4=36; z^2=4.4=16$

Kết hợp với đkxđ suy ra:
$(x,y,z)=(4,6,4); (-4; -6; -4)$

Em cảm ơn ạ

a) \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\) và \(x^2-y^2+2z^2=108\)

Giải

Áp dụng tính chất của dãy tỉ số bằng nhau:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x^2-y^2+2z^2}{2^2-3^2+2.4^2}=\dfrac{108}{27}=4\)

\(\dfrac{x}{2}=4\Rightarrow x=4.2=8\)

\(\dfrac{y}{3}=4\Rightarrow y=4.3=12\)

\(\dfrac{z}{4}=4\Rightarrow z=4.4=16\).

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Dựa vào t/c dãy tỉ số bằng nhau ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x^2-y^2+2z^2}{2^2-3^2+2.4^2}=\dfrac{108}{27}=4\)

\(x=2.4=8\)

\(y=3.4=12\)

\(z=4.4=16\)