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1.thay x=25 vào biểu thức A ta có:
25^3-15.25^2+75.25=8125
2.
a,x^3-3^3-x(x^2-2^2)-1=0
x^3-27-x^3+4x-1=0
4x-28=0
4(x-7)=0
X=7
b,(x^3+3x^2+3x+1)-(x^3-3x^2+3x-1)-6(x^2-2x+1)+10=0
x^3+3x^2+3x+1-X^3+3x^2-3x+1-6x^2+12x-6+10=0
12x+6=0
6(2x+1)=0
2x+1=0
2x=-1
x=-1/2
**** cho mk nha!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(\left(10x+3\right):8=\left(7-8x\right):12\)
\(\left(10x+3\right).\frac{1}{8}=\left(7-8x\right).\frac{1}{12}\)
\(\frac{5}{4}x+\frac{3}{8}=\frac{7}{12}-\frac{8}{12}x\)
\(\frac{5}{4}x+\frac{8}{12}x=\frac{7}{12}-\frac{3}{8}\)
\(\frac{23}{12}x=\frac{5}{24}\)
\(x=\frac{5}{46}\)
E mới lớp 6 nên giải sai thì thông cảm ạ UwU
\(b,\frac{x}{10}-\left(\frac{x}{30}+\frac{2x}{45}\right)=\frac{4}{5}\)
\(< =>\frac{9x}{90}-\frac{7x}{90}=\frac{4}{5}\)
\(< =>\frac{x}{45}=\frac{32}{45}\)
\(< =>x=32\)
\(d,\frac{10x+3}{8}=\frac{7-8x}{12}\)
\(< =>\left(10x+3\right).12=\left(7-8x\right).8\)
\(< =>120x+36=56-64x\)
\(< =>184x=56-36=20\)
\(< =>x=\frac{20}{184}=\frac{5}{46}\)
Câu d : \({2x \over x+1}\) + \({18\over x^2+2x-3}\) = \({2x-5 \over x+3}\)
a) \(x^4+2x^3-3x^2-8x-4=0\)
\(\Leftrightarrow x^4+2x^3-3x^2-6x-2x-4=0\)
\(\Leftrightarrow x^3\left(x+2\right)-3x\left(x+2\right)-2\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^3-3x-2=0\right)\)
\(\Leftrightarrow\left(x+2\right)\left(x^3-4x+x-2=0\right)\)
\(\Leftrightarrow\left(x+2\right)\left[x\left(x^2-4\right)+\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left[x\left(x-2\right)\left(x+2\right)+\left(x-2\right)\right]=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-2\right)\left(x+1\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\pm2\\x=-1\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{\pm2;-1\right\}\)
b) \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=0\)
\(\Leftrightarrow x-2=0\)hoặc \(x+2=0\)hoặc \(x^2-10=0\)
\(\Leftrightarrow x=2\)hoặc \(x=-2\)hoặc \(x=\pm\sqrt{10}\)
Vậy tập nghiệm của phương trình là : \(S=\left\{\pm2;\pm\sqrt{10}\right\}\)
c) \(2x^3+7x^2+7x+2=0\)
\(\Leftrightarrow2x^3+2x^2+5x^2+5x+2x+2=0\)
\(\Leftrightarrow2x^2\left(x+1\right)+5x\left(x+1\right)+2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x^2+5x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\2x^2+5x+2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\left(tm\right)\\2\left(x+\frac{5}{4}\right)^2+\frac{7}{16}=0\left(ktm\right)\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-1\right\}\)
d) Xem lại đề
1) \(2x.\left(x-7\right)-\left(x+3\right)\left(x-2\right)-\left(x+4\right)\left(x-4\right)\)
\(=2x^2-14x-\left(x^2+x-6\right)-\left(x^2-4\right)\)
\(=-15x+10\)
b) \(2x.\left(x+1\right)^2-\left(x-1\right)^3-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=2x.\left(x^2+2x+1\right)-\left(x^3-3x^2+3x-1\right)-\left(x^3-8\right)\)
\(=2x^3+4x^2+2x-x^3+3x^2-3x+1-x^3+8\)
\(=7x^2-x+9\)
c) \(\left(x-5\right)\left(x+5\right)\left(x+2\right)-\left(x+2\right)^3\)
\(=\left(x+2\right).\left[\left(x-5\right)\left(x+5\right)-\left(x+2\right)^2\right]\)
\(=\left(x+2\right).\left(x^2-25-x^2-4x-4\right)\)
\(=\left(x+2\right)\left(-4x-29\right)\)
\(=-4x^2-37x-58\)
d) \(\left(x-3\right)^3+\left(x-5\right)\left(x^2+5x+25\right)-\left(x-1\right)\left(x^2+x+1\right)\)
\(=x^3-9x^2+27x-27+\left(x^3-125\right)-\left(x^3-1\right)\)
\(=x^3-9x^2+27x-151\)
e) \(\left(x-1\right)^3-\left(x-2\right)\left(x^2-2x+4\right)+3x^2+2x\)
\(=x^3-3x^2+3x-1-\left(x^3-8\right)+3x^2+2x\)
\(=5x+7\)
Nhẩm ấy, ko nháp âu
\(2x\left(x-7\right)-\left(x+3\right)\left(x-2\right)-\left(x+4\right)\left(x-4\right)\)
\(=2x^2-14x-\left(x^2-2x+3x-6\right)-\left(x^2-4x+4x-16\right)\)
\(=2x^2-14x-x^2+x-6-x^2+16\)
\(=-13x-10\)
\(2x\left(x+1\right)^2-\left(x-1\right)^3-\left(x-2\right)\left(x^2+2x+4\right)\)
\(=2x\left(x^2+2x+1\right)-\left(x^3-3x^2+3x-1\right)-\left(x-2\right)\left(x+2\right)\)
\(-2x^3+4x^2+2x-x^3+3x^2-3x+1-x^2+4\)
\(=-3x^3+6x^2-x+5\)
Mình cần gấp ạk
A và B