\(\text{∘ Ans}\)

\(\downarrow\)

`1,`

`86.15 + 150. 1,4`

`= 86. 15 + 15. 14`

`= 15.(86 + 14)`

`= 15.100`

`= 1500`

`2,`

`93.32 + 14.16`

`= 93.32 + 2.7.16`

`= 93.32 + 32.7`

`= 32.(93 + 7)`

`= 32.100`

`= 3200`

`3,`

\(98,6\cdot199-990\cdot9,86\)

`=`\(98,6\cdot199-99\cdot98,6\)

`=`\(98,6\cdot\left(199-99\right)\)

`=`\(98,6\cdot100\)

`=`\(9860\)

`4,`

\(85\cdot12,7+5\cdot3\cdot12,7?\)

`=`\(85\cdot12,7+15+12,7\)

`=`\(12,7\cdot\left(85+15\right)\)

`=`\(12,7\cdot100\)

`= 1270`

`5,`

\(0,12\cdot90-110\cdot0,6+36-25\cdot6?\)

\(=6\cdot1,8-11\cdot6+6\cdot6-25\cdot6\)

\(=6\cdot\left(1,8-11+6-25\right)\)

\(=6\cdot\left(-28,2\right)=-169,2\)

Phong

Xem lại bài làm và cấu trúc bài của lớp.

\(\text{∘ Ans}\)

\(\downarrow\)

\(\left(3-x\right)\cdot\left(4-x\right)\cdot\left(5-x\right)=0\)

`TH1:`

`3 - x = 0`

`\Rightarrow x = 3-0`

`\Rightarrow x = 3`

`TH2:`

`4 - x = 0`

`\Rightarrow x = 4 - 0`

`\Rightarrow x = 4`

`TH3:`

`5 - x = 0`

`=> x = 5 - 0`

`=> x = 5`

Vậy, `x = {3; 4; 5}.`

\(\text{∘ Ans}\)

\(\downarrow\)

\(\left(x^2-5x+4\right)\left(2x+4\right)-\left(2x^2-x-10\right)\left(x-3\right)\)

`= 2x(x^2 - 5x + 4) + 4(x^2 - 5x + 4) - [x(2x^2 - x - 10) - 3(2x^2 - x - 10) ]`

`= 2x^3 - 10x^2 + 8x + 4x^2 - 20x + 16 - (2x^3 - x^2 - 10x - 6x^2 + 3x + 30)`

`= 2x^3 - 6x^2 - 12x + 16 - 2x^3 + x^2 + 10x + 6x^2 - 3x - 30`

`= (2x^3 - 2x^3) + (-6x^2 + 6x^2 + x^2) + (-12x + 10x - 3x) + (16 - 30)`

`= x^2 - 5x - 14`

Bạn xem lại đề.

`@` `\text {Ans}`

`\downarrow`

\(6\times\left(5\times x+35\right)=330\)

`\Rightarrow 5 \times x + 35 = 330 \div 6`

`\Rightarrow 5 \times x + 35 = 55`

`\Rightarrow 5 \times x = 55 - 35`

`\Rightarrow 5 \times x = 20`

`\Rightarrow x = 20 \div 5`

`\Rightarrow x=4`

Vậy, `x=4.`

Có phải 31 đâu a? \(-3\dfrac{1}{2}\) mà c?

\(\text{∘}\) \(\text{Ans}\)

\(\downarrow\)

\(14x^2y^3-7xy^2\cdot\left(2x-3y\right)\)

`=`\(14x^2y^3-\left[7xy^2\cdot2x+7xy^2\cdot\left(-3y\right)\right]\)

`=`\(14x^2y^3-\left(14x^2y^2-21xy^3\right)\)

`=`\(14x^2y^3-14x^2y^2+21xy^3\)

\(\text{∘}\) \(\text{Kaizuu lv uuu.}\)

\(\dfrac{4}{9}-x=\dfrac{1}{8}\)

`\Rightarrow`\(x=\dfrac{4}{9}-\dfrac{1}{8}\)

`\Rightarrow`\(x=\dfrac{23}{72}\)

\(3\cdot\left[\left(4-x\right)3+51\right]\div3-2^2=14\)

\(\Rightarrow3\cdot\left(12-3x+51\right)\div3-4=14\)

\(\Rightarrow3\cdot\left(-3x+12+51\right)\div3=14+4\)

\(\Rightarrow3\cdot\left(-3x+63\right)\div3=18\)

\(\Rightarrow-9x+189=18\cdot3\)

\(\Rightarrow-9x+189=54\)

\(\Rightarrow-9x=54-189\)

\(\Rightarrow-9x=-135\)

\(\Rightarrow x=15\)

`2x(x-2)=0`

`\rightarrow 2x^2-4x=0`

`\rightarrow x(2x-4)=0`

`\rightarrow `\(\left[{}\begin{matrix}x=0\\2x-4=0\end{matrix}\right.\)

`\rightarrow `\(\left[{}\begin{matrix}x=0\\2x=4\end{matrix}\right.\)

`\rightarrow `\(\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

Vậy, nghiệm của đa thức là `x={0; 2}`

`-x^2+5x-4=0`

`\rightarrow x^2-5x+4=0`

`\rightarrow x^2-4x-x+4=0`

`\rightarrow (x^2-4x)-(x-4)=0`

`\rightarrow x(x-4)-(x-4)=0`

`\rightarrow (x-4)(x-1)=0`

`\rightarrow `\(\left[{}\begin{matrix}x-4=0\\x-1=0\end{matrix}\right.\)

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

Vậy, nghiệm của đa thức là `x={4; 1}.`