kh hiểu bn ơi

`4x=2+xx+1x<=>4x=2+3x<=>4x-3x=2<=>1x=2<=>x=2`

a: \(x^3-9x^2+6x+16\)

\(=x^3-8x^2-x^2+8x-2x+16\)

\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)\)

\(=\left(x-8\right)\left(x^2-x-2\right)\)

\(=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)

b: \(x^3-x^2-x-2\)

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

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

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

c: \(x^3+x^2-x+2\)

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

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

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

d: \(x^3-6x^2-x+30\)

\(=x^3+2x^2-8x^2-16x+15x+30\)

\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)

\(=\left(x+2\right)\left(x^2-8x+15\right)\)

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

e: Sửa đề: \(x^3-7x-6\)

\(=x^3-x-6x-6\)

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

\(=x\left(x-1\right)\left(x+1\right)-6\left(x+1\right)\)

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

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

f: \(27x^3-27x^2+18x-4\)

\(=27x^3-9x^2-18x^2+6x+12x-4\)

\(=9x^2\left(3x-1\right)-6x\left(3x-1\right)+4\left(3x-1\right)\)

\(=\left(3x-1\right)\left(9x^2-6x+4\right)\)

g: \(2x^3-x^2+5x+3\)

\(=2x^3+x^2-2x^2-x+6x+3\)

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

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

h: \(\left(x^2-3\right)^2+16\)

\(=x^4-6x^2+9+16\)

\(=x^4-6x^2+25\)

\(=x^4+10x^2+25-16x^2\)

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

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

g= 35x²+15x-49x-21-35x²-28x+10x-8

g=-52x-(-29)

tại x=-3 ta có 

g= -52.(-3).(-29)

g=4542

\(G=\left(5x-7\right)\left(7x+3\right)-\left(7x+2\right)\left(5x-4\right)\)

\(=35x^2+15x-49x-21-35x^2+28x-10x+8\)

\(=-16x-13\)

Thay x = -3 \(\Rightarrow G=35\)

Vậy G = 35 khi x = -3

\(G=\left(5x-7\right)\left(7x+3\right)-\left(7x+2\right)\left(5x-4\right)\)

\(=\left(35x^2+15x-49x-21\right)-\left(35x^2-28x+10x-8\right)\)

\(=35x^2+15x-49x-21-35x^2+28x-10x+8\)

\(=-16x-13\)

Thay \(=-3\) vào biểu thức \(G\) , ta được :

\(G=-16x-13=-16\left(-3\right)-13=48-13=35\)

\(G=\left(5x-7\right)\left(7x+3\right)-\left(7x+2\right)\left(5x-4\right)\)

\(G=35x^2+15x-49x-21-\left(35x^2-28x+10x-8\right)\)

\(G=35x^2-34x-21-35x^2+18x+8\)

\(G=-16x-13\)(1)

Thay \(x=-3\) vào (1) ta được:

\(-16.\left(-3\right)-13=48-13=35\)

Vậy..............

Chúc bạn hcọ tốt!!!

\(7x^2\left(x^2-5x+2\right)-5x\left(x^3-7x^2+3x\right)\)

\(=7x^4-35x^3+14x^2-5x^4+35x^3-15x^2\)

\(=2x^4-x^2\)

Thay: \(x=-\frac{1}{2}\) vào được

\(2.\left(-\frac{1}{2}\right)^4-\left(-\frac{1}{2}\right)^2\)

\(=2.\frac{1}{16}-\frac{1}{4}\)

\(=\frac{1}{8}-\frac{1}{4}=\frac{1}{8}-\frac{2}{8}=-\frac{1}{8}\)

P/s: Ko chắc

Câu 1: 

Ta có: \(x^2-5x+4=0\)

\(\Leftrightarrow x^2-x-4x+4=0\)

\(\Leftrightarrow\left(x-1\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\end{matrix}\right.\)

Vậy: S={1;4}

Câu 2: 

Ta có: \(3x^2-7x+3=0\)

\(\Delta=\left(-7\right)^2-4\cdot3\cdot3=49-36=13\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{7-\sqrt{13}}{6}\\x_2=\dfrac{7+\sqrt{13}}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{7-\sqrt{13}}{6};\dfrac{7+\sqrt{13}}{6}\right\}\)

Câu 3: 

Ta có: \(5x^2-x-4=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{4}{5}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{4}{5}\right\}\)

Câu 4: 

Ta có: \(7x^2+x-8=0\)

\(\Leftrightarrow\left(x-1\right)\left(7x+8\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{8}{7}\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{8}{7}\right\}\)

Câu 1x^2-5x+4=0

<=>(x-1)(x-4)=0

<=>[x=1;x=4

Câu 2 3x^2-7x+3=0

x=7/6-căn bậc hai(13)/6, x=căn bậc hai(13)/6+7/6

x=7/6-căn bậc hai(13)/6, x=căn bậc hai(13)/6+7/6

Câu 3 5*x^2 -x-4 = 0

x=-4/5, x=1

Câu 4 7*x^2 +x-8 = 0

x=-8/7, x=1

bn ơi mk giải thế có chỗ nào ko hiểu bn có thể hỏi mk nhé

Cứ thay vào rùi thính thui

Mấy bài kia phá tung tóe rồi rút gọn hết sức xong thay x vào, làm câu c thôi nhé:

c) \(C=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)

riêng câu này ta thay x = 9 vào luôn, vậy ta có:

\(C=9^{14}-10\cdot9^{13}+10\cdot9^{12}-10\cdot9^{11}+...+10\cdot9^2-10\cdot9+10\)

\(=9^{14}-\left(9+1\right)\cdot9^{13}+\left(9+1\right)\cdot9^{12}-\left(9+1\right)\cdot9^{11}+...+\left(9+1\right)\cdot9^2-\left(9+1\right)\cdot9+10\)

\(=9^{14}-9^{14}-9^{13}+9^{13}+9^{12}-9^{12}-9^{11}+...+9^3+9^2-9^2-9+10\)

\(=-9+10\)

\(=1\)