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\(b,\frac{x-3}{x-2}=\frac{5}{\left(x-2\right)\left(x+3\right)}\)ĐKXĐ : \(x\ne2;\ne-3\)
\(\Leftrightarrow\frac{x^2-9}{\left(x-2\right)\left(x+3\right)}=\frac{5}{\left(x-2\right)\left(x+3\right)}\)
\(\Leftrightarrow x^2-9=5\)
\(\Leftrightarrow x^2=14\)
\(x=\sqrt{14}\)
.....
a) \(\left(x+3\right)^2-\left(x-3\right)^2=6x\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2-6x+9\right)=6x\)
\(\Leftrightarrow x^2+6x+9-x^2+6x-9=6x\Leftrightarrow12x=6x\)\(\Leftrightarrow12x-6x=0\Leftrightarrow6x=0\Leftrightarrow x=0\)
Vậy phương trình có tập nghiệm S = { 0 }
b)\(-ĐKXĐ:\hept{\begin{cases}x-2\ne0\\\left(x-2\right)\left(x+3\right)\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne-3\end{cases}}\)
- Ta có : \(\frac{x-3}{x-2}=\frac{5}{\left(x-2\right)\left(x+3\right)}\Leftrightarrow\frac{x-3}{x-2}-\frac{5}{\left(x-2\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x+3\right)-5}{\left(x-2\right)\left(x+3\right)}=0\Leftrightarrow\left(x-3\right)\left(x+3\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\left(thoaman\right)\\x=-3\left(kothoaman\right)\end{cases}}\)
Vậy phương trình có tập nghiệm S = { 3 }
\(\Leftrightarrow\frac{6x+5}{12x+9}-\frac{3x-7}{12x-9}=\frac{4x^2+10x-7}{16x^2-9}.\)
\(\Leftrightarrow\frac{\left(6x+5\right)\left(12x-9\right)-\left(3x-7\right)\left(12x+9\right)}{\left(3.4.x\right)^2-\left(3.3\right)^2}=\frac{4x^2+10x-7}{16x^2-9}\)
\(\Leftrightarrow\frac{72x^2+6x-45-\left(36x^2-57x-63\right)}{3^2\left(16x^2-9\right)}=\frac{4x^2+10x-7}{16x^2-9}\)
ĐK: \(16x^2-9\ne0\Leftrightarrow x^2\ne\left(\frac{3}{4}\right)^2\Rightarrow x\ne\pm\frac{3}{4}\)
\(\Leftrightarrow72x^2+6x-45-36x^2+57x+63=36x^2+90x-63\)
\(\Leftrightarrow27x=81\Leftrightarrow x=3\)
chuyển vế sang r phân tích thành nhân tử, có thể dùng máy tính bỏ túi nhé bạn
câu 1: 9\(x^2\) + 12\(x\) + 5 =11
(3\(x\))2 + 2.3.\(x\) .2 + 22 + 1 = 11
(3\(x\) + 2)2 = 11 - 1
(3\(x\) + 2)2 = 10
\(\left[{}\begin{matrix}3x+2=\sqrt{10}\\3x+2=-\sqrt{10}\end{matrix}\right.\)
\(\left[{}\begin{matrix}3x=\sqrt{10}-2\\3x=-\sqrt{10}-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{\sqrt{10}-2}{3}\\x=\dfrac{-\sqrt{10}-2}{3}\end{matrix}\right.\)
Vậy S = {\(\dfrac{-\sqrt{10}-2}{3}\); \(\dfrac{\sqrt{10}-2}{3}\)}
Câu 2: 6\(x^2\) + 16\(x\) + 12 = 2\(x^2\)
6\(x^2\) + 16\(x\) + 12 - 2\(x^2\) = 0
4\(x^2\) + 16\(x\) + 12 = 0
(2\(x\))2 + 2.2.\(x\).4 + 16 - 4 = 0
(2\(x\) + 4)2 = 4
\(\left[{}\begin{matrix}2x+4=2\\2x+4=-2\end{matrix}\right.\)
\(\left[{}\begin{matrix}2x=-2\\2x=-6\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=-1\\x=-3\end{matrix}\right.\)
S = { -3; -1}
3, 16\(x^2\) + 22\(x\) + 11 = 6\(x\) + 5
16\(x^2\) + 22\(x\) - 6\(x\) + 11 - 5 = 0
16\(x^2\) + 16\(x\) + 6 = 0
(4\(x\))2 + 2.4.\(x\) . 2 + 22 + 2 = 0
(4\(x\) + 2)2 + 2 = 0 (1)
Vì (4\(x\)+ 2)2 ≥ 0 ∀
⇒ (4\(x\) + 2)2 + 2 > 0 ∀ \(x\) vậy (1) Vô nghiệm
S = \(\varnothing\)
Câu 4. 12\(x^2\) + 20\(x\) + 10 = 3\(x^2\) - 4\(x\)
12\(x^2\) + 20\(x\) + 10 - 3\(x^2\) + 4\(x\) = 0
9\(x^2\) + 24\(x\) + 10 = 0
(3\(x\))2 + 2.3.\(x\).4 + 16 - 6 = 0
(3\(x\) + 4)2 = 6
\(\left[{}\begin{matrix}3x+4=\sqrt{6}\\3x+4=-\sqrt{6}\end{matrix}\right.\)
\(\left[{}\begin{matrix}3x=-4+\sqrt{6}\\3x=-4-\sqrt{6}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=\dfrac{\sqrt{6}-4}{3}\\x=-\dfrac{\sqrt{6}+4}{3}\end{matrix}\right.\)
S = {\(\dfrac{-\sqrt{6}-4}{3}\); \(\dfrac{\sqrt{6}-4}{3}\)}
Lời giải:
a) $(x+3)^2-(x-3)^2=6x+18$
$\Leftrightarrow 12x=6x+18\Leftrightarrow 6x=18\Rightarrow x=3$
b) ĐK:$x\neq 2; x\neq 3$
PT $\Rightarrow x+3=\frac{5}{3-x}$
$\Rightarrow (x+3)(3-x)=5$
$\Rightarrow 9-x^2=5$
$\Rightarrow x^2=4\Rightarrow x=\pm 2$. Kết hợp với ĐKXĐ suy ra $x=-2$
c) ĐKXĐ: $x\neq \frac{\pm 3}{4}$
PT $\Leftrightarrow \frac{12x^2+30x-21}{(4x-3)(4x+3)}-\frac{(3x-7)(3x+4)}{(4x-3)(4x+3)}=\frac{(6x+5)(4x-3)}{(4x-3)(4x+3)}$
$\Rightarrow 12x^2+30x-21-(3x-7)(4x+3)=(6x+5)(4x-3)$
$\Leftrightarrow -24x^2+47x+15=0$
$\Rightarrow x=\frac{47\pm \sqrt{3649}}{48}$
d)
ĐK: $x\neq -1; x\neq 2$
PT $\Leftrightarrow \frac{4(x-2)}{(x+1)(x-2)}-\frac{2(x+1)}{(x-2)(x+1)}=\frac{x+3}{(x+1)(x-2)}$
$\Rightarrow 4(x-2)-2(x+1)=x+3$
$\Rightarrow x=13$ (t.m)
Mình ko ghi lại đề , bạn ghi ra xong rồi suy ra như mình nha .
1) \(=>A=\left(6x^2+3x-10x-5\right)-\left(6x^2+14x-9x-21\right)\)
\(=>A=-12x+16\)
2) \(=>B=8x^3+27-8x^3+2=29\)
3)\(=>C=[\left(x-1\right)-\left(x+1\right)]^3=\left(-2\right)^3=-8\)
4)\(=>D=[\left(2x+5\right)-\left(2x\right)]^3=5^3=125\)
5)\(=>E=\left(3x+1\right)^2-\left(3x+5\right)^2+12x+2\left(6x+3\right)\)
\(=>E=\left(3x+1+3x+5\right)\left(3x+1-3x-5\right)+12x+12x+6\)
\(=>E=\left(6x+6\right)\left(-4\right)+24x+6=-24x-24+24x+6=-18\)
6)\(=>F=\left(2x^2+3x-10x-15\right)-\left(2x^2-6x\right)+x+7=-8\)
k cho mik nha ,
\(A=x^2+9x+25\)
\(=x^2+2x\frac{9}{2}+\frac{81}{4}+\frac{19}{4}\)
\(=\left(x+\frac{9}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}\forall x\)
Dấu"="xảy ra khi \(\left(x+\frac{9}{2}\right)^2=0\Rightarrow x=\frac{-9}{2}\)
Vậy \(Min_A=\frac{19}{4}\Leftrightarrow x=\frac{-9}{2}\)
b,\(B=4x^2-8x+\frac{21}{2}\)
\(=4\left(x^2-2x+1\right)+\frac{13}{2}\)
\(=4\left(x-1\right)^2+\frac{13}{2}\ge\frac{13}{2}\forall x\)
Dấu"="xảy ra khi \(4\left(x-1\right)^2=0\Rightarrow x=1\)
Vậy \(Min_B=\frac{13}{2}\Leftrightarrow x=1\)
c,\(C=-x^2+2x+\frac{5}{2}\)
\(=-\left(x^2-2x-\frac{5}{2}\right)\)
\(=-\left(x^2-2x+1\right)+\frac{7}{2}\)
\(=-\left(x-1\right)^2+\frac{7}{2}\le\frac{7}{2}\forall x\)
Dấu"="xảy ra khi \(-\left(x-1\right)^2=0\Rightarrow x=1\)
Vậy\(Max_C=\frac{7}{2}\Leftrightarrow x=1\)
Bài 1.
A = x2 + 9x + 25
= ( x2 + 9x + 81/4 ) + 19/4
= ( x + 9/2 )2 + 19/4 ≥ 19/4 ∀ x
Đẳng thức xảy ra <=> x + 9/2 = 0 => x = -9/2
=> MinA = 19/4 <=> x = -9/2
B = 4x2 - 8x + 21/2
= 4( x2 - 2x + 1 ) + 13/2
= 4( x - 1 )2 + 13/2 ≥ 13/2 ∀ x
Đẳng thức xảy ra <=> x - 1 = 0 => x = 1
=> MinB = 13/2 <=> x = 1
C = -x2 + 2x + 5/2
= -( x2 - 2x + 1 ) + 7/2
= -( x - 1 )2 + 7/2 ≤ 7/2 ∀ x
Đẳng thức xảy ra <=> x - 1 = 0 => x = 1
=> MaxC = 7/2 <=> x = 1
D = -9x2 - 12x + 27/2
= -9( x2 + 4/3x + 4/9 ) + 35/2
= -9( x + 2/3 )2 + 35/2 ≤ 35/2 ∀ x
Đẳng thức xảy ra <=> x + 2/3 = 0 => x = -2/3
=> MaxD = 35/2 <=> x = -2/3
Bài 2.
a) 4x2 + 9y2 + 12x + 12y + 13 = 0
<=> ( 4x2 + 12x + 9 ) + ( 9y2 + 12y + 4 ) = 0
<=> ( 2x + 3 )2 + ( 3y + 2 )2 = 0 (*)
\(\hept{\begin{cases}\left(2x+3\right)^2\ge0\forall x\\\left(3y+2\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(2x+3\right)^2+\left(3y+2\right)^2\ge0\forall x,y\)
Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}2x+3=0\\3y+2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=-\frac{2}{3}\end{cases}}\)
=> x = -3/2 ; y = -2/3
b) 16x2 + 4y2 - 8x + 12y + 10 = 0
<=> ( 16x2 - 8x + 1 ) + ( 4y2 + 12y + 9 ) = 0
<=> ( 4x - 1 )2 + ( 2y + 3 )2 = 0 (*)
\(\hept{\begin{cases}\left(4x-1\right)^2\ge0\forall x\\\left(2y+3\right)^2\ge0\forall y\end{cases}}\Rightarrow\left(4x-1\right)^2+\left(2y+3\right)^2\ge0\forall x,y\)
Đẳng thức xảy ra ( tức (*) ) <=> \(\hept{\begin{cases}4x-1=0\\2y+3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{4}\\y=-\frac{3}{2}\end{cases}}\)
=> x = 1/4 ; y = -3/2
\(\frac{12x^2+30x-21}{16x^2-9}-\frac{3x-7}{3-4x}=\frac{6x+5}{4x+3}\)
ĐKXĐ: \(x\ne\pm\frac{3}{4}\)
\(< =>\frac{12x^2+30x-21}{\left(4x-3\right)\left(4x+3\right)}+\frac{3x-7}{4x-3}=\frac{6x+5}{4x+3}\)
\(=>12x^2+30x-21+\left(3x-7\right)\left(4x+3\right)=\left(6x+5\right)\left(4x-3\right)\)
\(< =>12x^2+30x-21+12x^2-19x-21=24x^2+2x-15\)
\(< =>24x^2+11x-42=24x^2+2x-15\)
\(< =>24x^2+11x-42-24x^2-2x+15=0\)
\(< =>9x-27=0\)
\(< =>x=3\left(TM\right)\)
Tập nghiệm phương trình \(S=\left\{3\right\}\)
\(\frac{12x^2+30x-21}{\left(4x-3\right)\left(4x+3\right)}\)-\(\frac{3x-7}{3-4x}\)=\(\frac{6x+5}{4x+3}\)
\(\frac{12x^2+30x-21}{\left(4x-3\right)\left(4x+3\right)}\)+\(\frac{\left(3x-7\right)\left(4x+3\right)}{\left(4x-3\right)\left(4x+3\right)}\)=\(\frac{\left(6x+5\right)\left(4x-3\right)}{\left(4x-3\right)\left(4x+3\right)}\)
12x2+30x-21+12x2-28x+9x-21=24x2+20x-18x-15
12x2+12x2-24x2+30x-28x+9x-20x+18x=21+21-15
-9x =27
x =\(\frac{27}{-9}\)
x =-3