Giúp tui vs mn

lớp 8 có pt bậc 2 ak??

Có nhưng giải bằng PT tích nhé

a) => 6x-6-5x=15

=> 6x-6+6-5x=15+6

=> 6x-5x=21

=> x=21

b) => 8x-4-7x=16

=> 8x-4+4-7x=16+4

=> 8x-7x=20

=> x=20

c) => 8-[-(2.4x)]-9x-21=31

=> 8-(-8x)-9x=31+21

=> 8-(-8x)-9x=52

=> (-8x)-9x=8-52

=> (-8x)+(-9x)=-44

=> -17x=-44

=> x=(-44):(-17)

=> x=

câu c chắc tớ làm sai

a) x = 21

b) x = 20

c) x = -4

a: Ta có: \(7x+25=144\)

\(\Leftrightarrow7x=119\)

hay x=17

b: Ta có: \(33-12x=9\)

\(\Leftrightarrow12x=24\)

hay x=2

c: Ta có: \(128-3\left(x+4\right)=23\)

\(\Leftrightarrow3\left(x+4\right)=105\)

\(\Leftrightarrow x+4=35\)

hay x=31

d: Ta có: \(71+\left(726-3x\right)\cdot5=2246\)

\(\Leftrightarrow5\left(726-3x\right)=2175\)

\(\Leftrightarrow726-3x=435\)

\(\Leftrightarrow3x=291\)

hay x=97

e: Ta có: \(720:\left[41-\left(2x+5\right)\right]=40\)

\(\Leftrightarrow41-\left(2x+5\right)=18\)

\(\Leftrightarrow2x+5=23\)

\(\Leftrightarrow2x=18\)

hay x=9

Bn cần bài nào vậy

1,ta có:h(x) = ( x - 3 ).( 16 - 4x )=0

*)x-3=0

=>x=3

*)16-4x=0

=>4x=16

=>x=4

2,ta có:4x^2 - 6x=0

<=>2x(2x-3)=0

*)2x=0

=>x=0

*)2x-3=0

=>2x=3

=>x=\(\frac{3}{2}\)

3,ta có:x^2 + 7x - 8=0

denta:7²-(-4(1.8))=81

x₁:(-7+9):2=1

x₂:(-7-81):2=-8

câu 2 you chả giống kqua tôi gì

h(x)=f(x)-g(x)= 5x^3+3x^2-9x+1-4x³+7x³+6x+16

=8x³+3x²-3x+17

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

                \(g\left(x\right)=4x^3-7x^3-6x-16\)

\(h\left(x\right)=f\left(x\right)-g\left(x\right)=\left(5x^3+3x^2-9x+1\right)-\left(4x^3-7x^3-6x-16\right)\)

                                 \(=5x^3+3x^2-9x+1-4x^3+7x^3+6x+16\)

                                 \(=8x^3+3x^2-3x+17\)