Ôn tập cuối năm phần số học - Hỏi đáp

\(a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b\right)\left(a+b-c\right)\)

a² + 2ab + b² - ac - bc

= ( a² + 2ab + b² ) - ac - bc

= (a + b ) ² - c ( a + b )

= ( a+ b ) ( a + b - c )

\(F=\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}-1+2}=\frac{\sqrt{a}+1}{\sqrt{a}}=1+\frac{1}{\sqrt{a}}\)

- Xin phép giải lại cho HOÀN CHỈNH

ĐKXĐ : \(\left\{{}\begin{matrix}a-\sqrt{a}\ne0\\a-1\ne0\\a\ge0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}\sqrt{a}\left(\sqrt{a}-1\right)\ne0\\\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)\ne0\\a\ge0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\sqrt{a}\ne0\\\sqrt{a}-1\ne0\\a\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}a\ne0\\a\ne1\\a\ge0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}a>0\\a\ne1\end{matrix}\right.\)

Ta có : \(F=\left(\frac{1}{\sqrt{a}-1}+\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\)

=> \(F=\left(\frac{1}{\sqrt{a}-1}+\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

=> \(F=\left(\frac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

=> \(F=\left(\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)

=> \(F=\left(\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{1}{\sqrt{a}-1}\right)\)

=> \(F=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}=\frac{\sqrt{a}+1}{\sqrt{a}}\)

Vậy ...

đkxd : x >=0 ; x khác 1

\(B=\frac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=\sqrt{x}+1+\sqrt{x}-2=2\sqrt{x}-1\)

d/ đkxd : x >= 0 ;x khác 9

\(D=\left(\frac{5}{x-\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}-3}=\frac{5+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)}\cdot\left(\sqrt{x}-3\right)=1\)

e/ đkxđ : a > 0 ; a khác 1

\(E=\frac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\frac{\sqrt{a}-1}{\sqrt{a}}=1-\frac{1}{\sqrt{a}}\)

a/ đkxđ : x >=0 ; x khác 1

\(A=1+\frac{3}{\sqrt{x}+1}-\frac{3}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\cdot\left(\sqrt{x}-1\right)=1\)

b/ đkxđ : x >=0 ; x khác 4

\(B=\left(\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-2}+6\right)\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-3\right)=\left(\sqrt{x}-2+6\right)\left(\sqrt{x}-1-3\right)=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)=x-16\)

c/ đkxđ : x >= 0 ; x khác 1

\(C=\frac{2x+\sqrt{x}-1-\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2x-2}{\sqrt{x}\left(x-1\right)}=\frac{2}{\sqrt{x}}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

Ta có :

\(A=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{x-1}:\frac{1}{\sqrt{x}-1}\)

\(=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}.\left(\sqrt{x}-1\right)\)

\(=\frac{\sqrt{x}+4}{\sqrt{x}+1}-\frac{3}{\sqrt{x}+1}\)

\(=\frac{\sqrt{x}+1}{\sqrt{x}+1}\)

\(=1\)

Vậy...

b/ ĐKXĐ : \(\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

Ta có :

\(B=\left(\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}+6\right)\left(\frac{x\sqrt{x}-1}{x+\sqrt{x}+1}-3\right)\)

\(=\left(\frac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}-2}+6\right)\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-3\right)\)

\(=\left(\sqrt{x}-2+6\right)\left(\sqrt{x}-1-3\right)\)

\(=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)\)

\(=x-16\)

Vậy..

c/ ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

Ta có :

\(C=\frac{2\sqrt{x}}{x-1}+\frac{1}{x+\sqrt{x}}+\frac{1}{\sqrt{x}-x}\)

\(=\frac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}-\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)

\(=\frac{2x}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{2x+\sqrt{x}-1-\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)

\(=\frac{2x-2}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{2\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{2}{\sqrt{x}}\)

Vậy..

a/ \(x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)\)

b/ \(\sqrt{x}-x=\sqrt{x}\left(1-\sqrt{x}\right)\)

c/ \(x-1=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

d/ \(x\sqrt{x}-1=\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)

f/ \(x-\sqrt{x}-6=x+2\sqrt{x}-3\sqrt{x}-6=\sqrt{x}\left(\sqrt{x}+2\right)-3\left(\sqrt{x}+2\right)=\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)\)

g/ \(a-2\sqrt{a}+1=\left(\sqrt{a}-1\right)^2\)

h/ \(x-4\sqrt{x}+4=\left(\sqrt{x}-2\right)^2\)

1) Ta có: \(\left(x^2-4x+4\right)\left(x^2+4x+4\right)-\left(7x+4\right)^2=0\)

\(\Leftrightarrow\left(x-2\right)^2\cdot\left(x+2\right)^2-\left(7x+4\right)^2=0\)

\(\Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\right]^2-\left(7x+4\right)^2=0\)

\(\Leftrightarrow\left(x^2-4\right)^2-\left(7x+4\right)^2=0\)

\(\Leftrightarrow\left(x^2-4-7x-4\right)\left(x^2-4+7x+4\right)=0\)

\(\Leftrightarrow\left(x^2-7x-8\right)\left(x^2+7x\right)=0\)

\(\Leftrightarrow x\left(x+7\right)\left(x^2-8x+x-8\right)=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\\x-8=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\\x=8\\x=-1\end{matrix}\right.\)

Vậy: S={0;-7;8;-1}

2) Ta có: \(x^3-8x^2+17x-10=0\)

\(\Leftrightarrow x^3-2x^2-6x^2+12x+5x-10=0\)

\(\Leftrightarrow x^2\left(x-2\right)-6x\left(x-2\right)+5\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-6x+5\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x^2-x-5x+5\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=5\end{matrix}\right.\)

Vậy: S={2;1;5}

3) Ta có: \(2x^3-5x^2-x+6=0\)

\(\Leftrightarrow2x^3-4x^2-x^2+2x-3x+6=0\)

\(\Leftrightarrow2x^2\left(x-2\right)-x\left(x-2\right)-3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x^2-x-3\right)=0\)

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

\(\Leftrightarrow\left(x-2\right)\left[x\left(2x-3\right)+\left(2x-3\right)\right]=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\2x=3\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\\x=-1\end{matrix}\right.\)

Vậy: \(S=\left\{2;\frac{3}{2};-1\right\}\)

4) Ta có: \(4x^4-4x^2-3=0\)

\(\Leftrightarrow4x^4-6x^2+2x^2-3=0\)

\(\Leftrightarrow2x^2\left(2x^2-3\right)+\left(2x^2-3\right)=0\)

\(\Leftrightarrow\left(2x^2-3\right)\left(2x^2+1\right)=0\)

mà \(2x^2+1>0\forall x\in R\)

nên \(2x^2-3=0\)

\(\Leftrightarrow2x^2=3\)

\(\Leftrightarrow x^2=\frac{3}{2}\)

hay \(x=\pm\sqrt{\frac{3}{2}}\)

Vậy: \(S=\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}}\right\}\)

\(\sqrt{3}+1=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{4+2\sqrt{3}}>\sqrt{\sqrt{3}+2\sqrt{3}}=\sqrt{3\sqrt{3}}\) (Do 4 > căn 3 )

Vậy .....

1: Ta có: \(\left|x^2-72\right|-2=70\)

\(\Leftrightarrow\left|x^2-72\right|=72\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-72=72\\x^2-72=-72\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=144\\x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-12\\x=0\end{matrix}\right.\)

Vậy: S={0;-12;12}

2: Ta có: \(x^2-5\left|x\right|=14\)

\(\Leftrightarrow5\left|x\right|=x^2-14\)(*)

Trường hợp 1: x≥0

(*)\(\Leftrightarrow5x=x^2-14\)

\(\Leftrightarrow x^2-14=5x\)

\(\Leftrightarrow x^2-5x-14=0\)

\(\Leftrightarrow x^2-7x+2x-14=0\)

\(\Leftrightarrow x\left(x-7\right)+2\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)

Trường hợp 2: x<0

(*)\(\Leftrightarrow-5x=x^2-14\)

\(\Leftrightarrow x^2+5x-14=0\)

\(\Leftrightarrow x^2+7x-2x-14=0\)

\(\Leftrightarrow x\left(x+7\right)-2\left(x+7\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-7\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)

Vậy: S={-7;7}

Thay abc = 1 vào biểu thức ta có

\(\frac{a.abc}{ab+abc.a+abc}+\frac{b}{bc+b.acb+abc}+\frac{c}{ac+c+1}\)

= \(\frac{a^2bc}{ab+a^2bc+abc}+\frac{b}{bc+ab^2c+abc}+\frac{c}{ac+c+1}\)

= \(\frac{a^2bc}{ab\left(ac+c+1\right)}+\frac{b}{b\left(ac+c+1\right)}+\frac{c}{ac+c+1}\)

= \(\frac{ac}{\left(ac+c+1\right)}+\frac{1}{\left(ac+c+1\right)}+\frac{c}{ac+c+1}\)

= \(\frac{ac+c+1}{ac+c+1}\)

= 1 (đpcm)

Nếu có gì không hiểu nhớ nt cho mình nha

\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)

\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{abc}{a\cdot abc+abc+ab}\)

\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{a+1+ab}\)

\(=\frac{ab+a+1}{ab+a+1}=1\)

- Các ĐKXĐ tự tìm dùm mình hen :)

Ta có : \(D=\left(\frac{5}{x-\sqrt{x}-6}+\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}-3}\)

=> \(D=\left(\frac{5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}+\frac{1}{\sqrt{x}+2}\right)\left(\sqrt{x}-3\right)\)

=> \(D=\left(\frac{5+\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\right)\left(\sqrt{x}-3\right)\)

=> \(D=\left(\frac{\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)}\right)\left(\sqrt{x}-3\right)\)

=> \(D=\left(\frac{1}{\sqrt{x}-3}\right)\left(\sqrt{x}-3\right)=1\)

Ta có : \(E=\left(\frac{1}{a-\sqrt{a}}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a+1}}{a-2\sqrt{a}+1}\)

=> \(E=\left(\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\frac{1}{\sqrt{a}-1}\right):\frac{\sqrt{a+1}}{\left(\sqrt{a}-1\right)^2}\)

=> \(E=\left(\frac{\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\frac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)

=> \(E=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}=\frac{\sqrt{a}-1}{\sqrt{a}}\)

( làm đến đây thôi câu còn lại bạn tự làm hen )

_{Ghét nhất mấy câu viết sai đề b, c sai rất nhiều bạn ới}

Mình nghĩ bạn nên xem lại đề ví dụ như \(\sqrt{x}+1\rightarrow\sqrt{x+1}\) ( chắc do bạn đánh nhanh quá nên nhầm nha .

a, \(=\sqrt{x}\left(\sqrt{x}+1\right)\)

b, \(=\sqrt{x}\left(\sqrt{x}-1\right)\)

c, \(=\sqrt{x}\left(1-\sqrt{x}\right)\)

d, \(=\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)

e, \(=\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)

f, \(=x+2\sqrt{x}-3\sqrt{x}-6=\sqrt{x}\left(\sqrt{x}+2\right)-3\left(\sqrt{x}+2\right)\)

\(=\left(\sqrt{x}+2\right)\left(\sqrt{x}-3\right)\)

g, \(=\left(\sqrt{a}-1\right)^2\)

h, \(=\left(\sqrt{x}-2\right)^2\)