ôn thi trong sgk á, đề lớp 4 k nâng cao lm đôu UwU'

tự đi mà biết mà tôi cung k biết vì tôi chưa thi

Bài 1:

1: \(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{15\sqrt{x}-11-3x-9\sqrt{x}+2\sqrt{x}+6-2x+2\sqrt{x}-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

2: Thay x=1/9 vào A, ta được:

\(A=\left(-5\cdot\sqrt{\frac19}+2\right):\left(\sqrt{\frac19}+3\right)=\left(-5\cdot\frac13+2\right):\left(\frac13+3\right)\)

\(=\frac13:\frac{10}{3}=\frac{1}{10}\)

Thay \(x=4-2\sqrt3=\left(\sqrt3-1\right)^2\) vào A, ta được:

\(A=\frac{-5\cdot\sqrt{\left(\sqrt3-1\right)^2}+2}{\sqrt{\left(\sqrt3-1\right)^2}+3}=\frac{-5\left(\sqrt3-1\right)+2}{\sqrt3-1+3}\)

\(=\frac{-5\sqrt3+7}{2+\sqrt3}=\left(7-5\sqrt3\right)\left(2-\sqrt3\right)=14-7\sqrt3-10\sqrt3+15=29-17\sqrt3\)

3: \(A=\frac12\)

=>\(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=\frac12\)

=>\(-10\sqrt{x}+4=\sqrt{x}+3\)

=>\(-11\sqrt{x}=3-4=-1\)

=>\(\sqrt{x}=\frac{1}{11}\)

=>\(x=\frac{1}{121}\) (nhận)

4: A>-1

=>A+1>0

=>\(\frac{-5\sqrt{x}+2+\sqrt{x}+3}{\sqrt{x}+3}>0\)

=>\(-4\sqrt{x}+5>0\)

=>\(-4\sqrt{x}>-5\)

=>\(\sqrt{x}<\frac54\)

=>0<=x<25/16

Kết hợp DKXĐ, ta được: 0<=x<25/16 và x<>1

5: A nguyên khi \(-5\sqrt{x}+2\) ⋮\(\sqrt{x}+3\)

=>\(-5\sqrt{x}-15+17\) ⋮\(\sqrt{x}+3\)

=>17⋮\(\sqrt{x}+3\)

mà \(\sqrt{x}+3\ge3\forall x\) thỏa mãn ĐKXĐ

nên \(\sqrt{x}+3=17\)

=>\(\sqrt{x}=14\)

=>x=196(nhận)

6: \(A+5=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}+5=\frac{-5\sqrt{x}+2+5\sqrt{x}+15}{\sqrt{x}+3}=\frac{17}{\sqrt{x}+3}>0\)

=>A>-5

Bài 2:

ĐKXĐ: x>=0; x<>4

1: \(B=\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{x+5}{x-\sqrt{x}-2}\)

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

\(=\frac{x-3\sqrt{x}+2-x-4\sqrt{x}-3-x-5}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}=\frac{-x-7\sqrt{x}-6}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

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

2: Khi x=9 thì \(B=\frac{-3-6}{3-2}=-9\)

Khi \(x=9-4\sqrt5=\left(\sqrt5-2\right)^2\) vào B, ta được:

\(B=\frac{-\sqrt{\left(\sqrt5-2\right)^2}-6}{\sqrt{\left(\sqrt5-2\right)^2}-2}=\frac{-\sqrt5+2-6}{\sqrt5-2-2}=\frac{-\sqrt5-4}{\sqrt5-4}\)

\(=\frac{4+\sqrt5}{4-\sqrt5}=\frac{\left(4+\sqrt5\right)^2}{\left(4-\sqrt5\right)\left(4+\sqrt5\right)}=\frac{16+8\sqrt5+5}{16-5}=\frac{21+8\sqrt5}{11}\)

3: B=-1

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

=>-6=2(loại)

4: B<-1

=>B+1<0

=>\(\frac{-\sqrt{x}-6+\sqrt{x}-2}{\sqrt{x}-2}<0\)

=>\(-\frac{8}{\sqrt{x}-2}<0\)

=>\(\sqrt{x}-2>0\)

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

=>x>4

`@` `\text {Ans}`

`\downarrow`

`a)`

\(\left(\dfrac{7}{8}-\dfrac{3}{4}\right)\cdot1\dfrac{1}{3}-\dfrac{2}{3}\cdot0,5\)

`=`\(\dfrac{1}{8}\cdot\dfrac{4}{3}-\dfrac{1}{3}\)

`=`\(\dfrac{1}{6}-\dfrac{1}{3}=-\dfrac{1}{6}\)

`b)`

\(\left(2+\dfrac{5}{6}\right)\div1\dfrac{1}{5}+\left(-\dfrac{7}{12}\right)\)

`=`\(\dfrac{17}{6}\div1\dfrac{1}{5}-\dfrac{7}{12}\)

`=`\(\dfrac{85}{36}-\dfrac{7}{12}=\dfrac{16}{9}\)

`c)`

\(75\%-1\dfrac{1}{2}+0,5\div\dfrac{5}{12}\)

`=`\(-\dfrac{3}{4}+\dfrac{6}{5}=\dfrac{9}{20}\)

a) \(\left(\dfrac{7}{8}-\dfrac{3}{4}\right).1\dfrac{1}{3}-\dfrac{2}{3}.0,5\)

\(=\left(\dfrac{7}{8}-\dfrac{6}{8}\right).\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)

\(=\dfrac{1}{8}.\dfrac{4}{3}-\dfrac{2}{3}.\dfrac{1}{2}\)

\(=\dfrac{1}{6}-\dfrac{1}{3}\)

\(=\dfrac{-1}{6}\)

b) \(\left(2+\dfrac{5}{6}\right):1\dfrac{1}{5}+\dfrac{-7}{12}\)

\(=\left(\dfrac{12}{6}+\dfrac{5}{6}\right):\dfrac{6}{5}+\dfrac{-7}{12}\)

\(=\dfrac{17}{6}.\dfrac{5}{6}+\dfrac{-7}{12}\)

\(=\dfrac{85}{36}+\dfrac{-7}{12}\)

\(=\dfrac{16}{9}\)

c) \(75\%-1\dfrac{1}{2}+0,5:\dfrac{5}{12}\)

\(=\dfrac{3}{4}-\dfrac{3}{2}+\dfrac{1}{2}.\dfrac{12}{5}\)

\(=\dfrac{3}{4}-\dfrac{6}{4}+\dfrac{6}{5}\)

\(=\dfrac{-3}{4}+\dfrac{6}{5}\)

\(=\dfrac{9}{20}\)

Đặt \(A=3x^2+y^2-2xy-6x-2y+11\)

\(=3x^2+y^2-2y\left(x+1\right)+\left(x+1\right)^2-\left(x+1\right)^2-6x+11\)

\(=\left(y-x-1\right)^2+3x^2-x^2-2x-1-6x+11\)

\(=\left(x-y+1\right)^2+2x^2-8x+10\)

\(=\left(x-y+1\right)^2+2\left(x^2-4x+5\right)\)

\(=\left(x-y+1\right)^2+2\left(x^2-4x+4+1\right)=\left(x-y+1\right)^2+2\left(x-2\right)^2+2\ge2\forall x,y\)

Dấu '=' xảy ra khi \(\begin{cases}x-y+1=0\\ x-2=0\end{cases}\Rightarrow\begin{cases}x=2\\ y=x+1=2+1=3\\ \end{cases}\)

a, 2\(xy\) - 2\(x\) + 3\(y\) = -9

(2\(xy\) - 2\(x\)) + 3\(y\) - 3 = -12

2\(x\)(\(y-1\)) + 3(\(y-1\)) = -12

(\(y-1\))(2\(x\) + 3) = -12

Ư(12) = {-12; -6; -4; -3; -2; -1; 1; 2; 3; 4; 6; 12}

Lập bảng ta có:

Theo bảng trên ta có: Các cặp \(x\);\(y\) nguyên thỏa mãn đề bài là:

(\(x;y\)) = (-1; -11); (0; -3); (-3; 5); ( -2; 13)

b, (\(x+1\))²(\(y\) - 3) = -4 

    Ư(4) = {-4; -2; -1; 1; 2; 4}

Lập bảng ta có: 

Theo bảng trên ta có: các cặp \(x;y\) nguyên thỏa mãn đề bài là: 

(\(x;y\)) = (0; -1); (-3; 2); (1; 2)

Sửa đề :

5327 x 3,5 + 532,7 x 6,4 x 10 = 53270 x 0,01 thành   

5327 x 3,5 + 532,7 x 6,4 x 10 + 53270 x 0,01 

= 532,7 x 35 + 532,7 x 64 + 532,7 

= 532,7 x ( 35 + 64 + 1 )

= 532,7 x 100

= 53270 ( mình làm hơi tắt , mong bạn thông cảm )

= 532,7 

hihi

#hok tốt#

KHO THE

\(A=\frac{\left[\left(25-1\right):1+1\right]\left(25+1\right)}{2}=325.\)

\(B=\frac{\left[\left(51-3\right):2+1\right]\left(51+3\right)}{2}=675\)

\(C=\frac{\left[\left(81-1\right):4+1\right]\left(81+1\right)}{2}=861\)

2. \(2xy-4y+2y^2\)

 \(=2y\left(x-2+y\right)\)